Codeforces Round #375 (Div. 2) A B C 水模拟贪心

There are three friend living on the straight line Ox in Lineland. The first friend lives at the point x₁, the second friend lives at the point x₂, and the third friend lives at the point x₃. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?

It's guaranteed that the optimal answer is always integer.

Input

The first line of the input contains three distinct integers x₁, x₂ and x₃ (1 ≤ x₁, x₂, x₃ ≤ 100) — the coordinates of the houses of the first, the second and the third friends respectively.

Output

Print one integer — the minimum total distance the friends need to travel in order to meet together.

Examples

Input

7 1 4

Output

Input

30 20 10

Output

Note

In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.

题意：

题解：排序输出a[2]-a[0]

 1 /******************************
 2 code by drizzle
 3 blog: www.cnblogs.com/hsd-/
 4 ^ ^    ^ ^
 5  O      O
 6 ******************************/
 7 #include<bits/stdc++.h>
 8 #include<map>
 9 #include<set>
10 #include<cmath>
11 #include<queue>
12 #include<bitset>
13 #include<math.h>
14 #include<vector>
15 #include<string>
16 #include<stdio.h>
17 #include<cstring>
18 #include<iostream>
19 #include<algorithm>
20 #pragma comment(linker, "/STACK:102400000,102400000")
21 using namespace std;
22 #define  A first
23 #define B second
24 const int mod=1000000007;
25 const int MOD1=1000000007;
26 const int MOD2=1000000009;
27 const double EPS=0.00000001;
28 //typedef long long ll;
29 typedef __int64 ll;
30 const ll MOD=1000000007;
31 const int INF=1000000010;
32 const ll MAX=1ll<<55;
33 const double eps=1e-5;
34 const double inf=~0u>>1;
35 const double pi=acos(-1.0);
36 typedef double db;
37 typedef unsigned int uint;
38 typedef unsigned long long ull;
39 int n;
40 int a[3];
41 int main()
42 {
43     scanf("%d %d %d",&a[0],&a[1],&a[2]);
44     sort(a,a+3);
45     cout<<a[1]-a[0]+a[2]-a[1]<<endl;
46     return 0;
47 }

Modern text editors usually show some information regarding the document being edited. For example, the number of words, the number of pages, or the number of characters.

In this problem you should implement the similar functionality.

You are given a string which only consists of:

uppercase and lowercase English letters,
underscore symbols (they are used as separators),
parentheses (both opening and closing).

It is guaranteed that each opening parenthesis has a succeeding closing parenthesis. Similarly, each closing parentheses has a preceding opening parentheses matching it. For each pair of matching parentheses there are no other parenthesis between them. In other words, each parenthesis in the string belongs to a matching "opening-closing" pair, and such pairs can't be nested.

For example, the following string is valid: "_Hello_Vasya(and_Petya)__bye_(and_OK)".

Word is a maximal sequence of consecutive letters, i.e. such sequence that the first character to the left and the first character to the right of it is an underscore, a parenthesis, or it just does not exist. For example, the string above consists of seven words: "Hello", "Vasya", "and", "Petya", "bye", "and" and "OK". Write a program that finds:

the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
the number of words inside the parentheses (print 0, if there is no word inside the parentheses).

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 255) — the length of the given string. The second line contains the string consisting of only lowercase and uppercase English letters, parentheses and underscore symbols.

Output

Print two space-separated integers:

the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
the number of words inside the parentheses (print 0, if there is no word inside the parentheses).

Examples

Input

37
_Hello_Vasya(and_Petya)__bye_(and_OK)

Output

5 4

Input

37
_a_(_b___c)__de_f(g_)__h__i(j_k_l)m__

Output

2 6

Input

27
(LoooonG)__shOrt__(LoooonG)

Output

5 2

Input

5
(___)

Output

0 0

Note

In the first sample, the words "Hello", "Vasya" and "bye" are outside any of the parentheses, and the words "and", "Petya", "and" and "OK" are inside. Note, that the word "and" is given twice and you should count it twice in the answer.

题意：统计不在括号内的单词的最大长度和括号内的单词数量

题解：模拟

  1 /******************************
  2 code by drizzle
  3 blog: www.cnblogs.com/hsd-/
  4 ^ ^    ^ ^
  5  O      O
  6 ******************************/
  7 #include<bits/stdc++.h>
  8 #include<map>
  9 #include<set>
 10 #include<cmath>
 11 #include<queue>
 12 #include<bitset>
 13 #include<math.h>
 14 #include<vector>
 15 #include<string>
 16 #include<stdio.h>
 17 #include<cstring>
 18 #include<iostream>
 19 #include<algorithm>
 20 #pragma comment(linker, "/STACK:102400000,102400000")
 21 using namespace std;
 22 #define  A first
 23 #define B second
 24 const int mod=1000000007;
 25 const int MOD1=1000000007;
 26 const int MOD2=1000000009;
 27 const double EPS=0.00000001;
 28 //typedef long long ll;
 29 typedef __int64 ll;
 30 const ll MOD=1000000007;
 31 const int INF=1000000010;
 32 const ll MAX=1ll<<55;
 33 const double eps=1e-5;
 34 const double inf=~0u>>1;
 35 const double pi=acos(-1.0);
 36 typedef double db;
 37 typedef unsigned int uint;
 38 typedef unsigned long long ull;
 39 int n;
 40 char a[300];
 41 char  b[300];
 42 int main()
 43 {
 44     scanf("%d",&n);
 45     scanf("%s",a);
 46     for(int i=0;i<n;i++)
 47         b[i]=a[i];
 48     for(int i=0;i<n;i++)
 49     {
 50         int j=i;
 51         if(a[i]=='(')
 52         {
 53             a[i]='_';
 54             while(1)
 55             {
 56                 j++;
 57                 if(a[j]==')')
 58                     break;
 59                 a[j]='_';
 60             }
 61             a[j]='_';
 62         }
 63         i=j;
 64     }
 65     for(int i=0;i<n;i++)
 66     {
 67       if(a[i]==b[i])
 68         b[i]='_';
 69       if(b[i]=='('||b[i]==')')
 70         b[i]='_';
 71     }
 72     int ans1=0,ans2=0;
 73     for(int i=0;i<n;i++)
 74     {
 75         int j=i;
 76         int exm=0;
 77         if(a[i]!='_')
 78         {
 79             exm++;
 80             j++;
 81             while(1)
 82             {
 83 
 84                 if(a[j]=='_'||j>=n)
 85                     break;
 86                 j++;
 87                 exm++;
 88             }
 89         }
 90         i=j;
 91         ans1=max(ans1,exm);
 92     }
 93     for(int i=0;i<n;i++)
 94     {
 95         int j=i;
 96         if(b[i]!='_')
 97         {
 98             j++;
 99             while(1)
100             {
101 
102                 if(b[j]=='_'||j>=n)
103                     break;
104                 j++;
105             }
106             ans2++;
107         }
108         i=j;
109     }
110     cout<<ans1<<" "<<ans2<<endl;
111     return 0;
112 }

Polycarp is a music editor at the radio station. He received a playlist for tomorrow, that can be represented as a sequence a₁, a₂, ..., a_n, where a_i is a band, which performs the i-th song. Polycarp likes bands with the numbers from 1 to m, but he doesn't really like others.

We define as b_j the number of songs the group j is going to perform tomorrow. Polycarp wants to change the playlist in such a way that the minimum among the numbers b₁, b₂, ..., b_m will be as large as possible.

Find this maximum possible value of the minimum among the b_j (1 ≤ j ≤ m), and the minimum number of changes in the playlist Polycarp needs to make to achieve it. One change in the playlist is a replacement of the performer of the i-th song with any other group.

Input

The first line of the input contains two integers n and m (1 ≤ m ≤ n ≤ 2000).

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹), where a_i is the performer of the i-th song.

Output

In the first line print two integers: the maximum possible value of the minimum among the b_j (1 ≤ j ≤ m), where b_j is the number of songs in the changed playlist performed by the j-th band, and the minimum number of changes in the playlist Polycarp needs to make.

In the second line print the changed playlist.

If there are multiple answers, print any of them.

Examples

Input

4 2
1 2 3 2

Output

2 1
1 2 1 2

Input

7 3
1 3 2 2 2 2 1

Output

2 1
1 3 3 2 2 2 1

Input

4 4
1000000000 100 7 1000000000

Output

1 4
1 2 3 4

Note

In the first sample, after Polycarp's changes the first band performs two songs (b₁ = 2), and the second band also performs two songs (b₂ = 2). Thus, the minimum of these values equals to 2. It is impossible to achieve a higher minimum value by any changes in the playlist.

In the second sample, after Polycarp's changes the first band performs two songs (b₁ = 2), the second band performs three songs (b₂ = 3), and the third band also performs two songs (b₃ = 2). Thus, the best minimum value is 2.

题意：歌单中有n首歌一个人喜欢喜欢1~m 乐队分别给出你n首歌的乐队要求歌单中来自喜欢的乐队的歌曲的数量的最小值尽量大现在可以更改歌单之中的歌输出这个最小值以及更改歌曲的次数（尽量少）以及更改之后的歌单

题解：贪心处理最小值为一个定值n/m 先处理m之后的乐队后处理1~m的乐队对于不需要更改的m之后的乐队直接输出

数据

4 3

1 2 3 4

answer

1 0

1 2 3 4

  1 /******************************
  2 code by drizzle
  3 blog: www.cnblogs.com/hsd-/
  4 ^ ^    ^ ^
  5  O      O
  6 ******************************/
  7 #include<bits/stdc++.h>
  8 #include<map>
  9 #include<set>
 10 #include<cmath>
 11 #include<queue>
 12 #include<bitset>
 13 #include<math.h>
 14 #include<vector>
 15 #include<string>
 16 #include<stdio.h>
 17 #include<cstring>
 18 #include<iostream>
 19 #include<algorithm>
 20 #pragma comment(linker, "/STACK:102400000,102400000")
 21 using namespace std;
 22 #define  A first
 23 #define B second
 24 const int mod=1000000007;
 25 const int MOD1=1000000007;
 26 const int MOD2=1000000009;
 27 const double EPS=0.00000001;
 28 //typedef long long ll;
 29 typedef __int64 ll;
 30 const ll MOD=1000000007;
 31 const int INF=1000000010;
 32 const ll MAX=1ll<<55;
 33 const double eps=1e-5;
 34 const double inf=~0u>>1;
 35 const double pi=acos(-1.0);
 36 typedef double db;
 37 typedef unsigned int uint;
 38 typedef unsigned long long ull;
 39 int n,m;
 40 int b[2005];
 41 int a[2005];
 42 int main()
 43 {
 44     scanf("%d %d",&n,&m);
 45     int ans1=0;
 46     int ans2=0;
 47     ans1=n/m;
 48     memset(b,0,sizeof(b));
 49     for(int i=1;i<=n;i++)
 50     {
 51       scanf("%d",&a[i]);
 52       if(a[i]<=m)
 53         {b[a[i]]++;}
 54     }
 55     for(int i=1;i<=n;i++)
 56     {
 57         if(a[i]>m)
 58         {
 59             for(int j=1;j<=m;j++)
 60             {
 61                 if(b[j]<ans1)
 62                 {
 63                     b[j]++;
 64                     a[i]=j;
 65                     ans2++;
 66                     break;
 67                 }
 68             }
 69 
 70         }
 71     }
 72     int flag=1;
 73     for(int i=1;i<=n;i++)
 74     {
 75          if(a[i]>m){
 76             flag=0;
 77             break;
 78         }
 79     }
 80     if(flag){
 81     for(int i=1;i<=n;i++)
 82     {
 83         if(b[a[i]]>ans1)
 84         {
 85             for(int j=1;j<=m;j++)
 86             {
 87                 if(b[j]<ans1)
 88                 {
 89                     b[a[i]]--;
 90                     a[i]=j;
 91                     b[j]++;
 92                     ans2++;
 93                     break;
 94                 }
 95             }
 96         }
 97     }
 98     }
 99     cout<<ans1<<" "<<ans2<<endl;
100     for(int i=1;i<=n;i++)
101         printf("%d ",a[i]);
102     printf("
");
103      return 0;
104 }

相关阅读:
Oracle数据库物理结构
 Oracle 性能调优之:使用 V$SQL_PLAN 视图查询内存中的执行计划
 Oracle的操作系统身份认证（转）
Oracle表空间知识
 Oracle数据库LOGGING&NOLOGGING模式概述
 使用插件bootstrap-table实现表格记录的查询、分页、排序等处理
 项目thymeleaf
Comparable和Comparator的区别
 AJAX
thymeleaf 基础
原文地址：https://www.cnblogs.com/hsd-/p/5929944.html

Codeforces Round #375 (Div. 2) A B C 水 模拟 贪心

Codeforces Round #375 (Div. 2) A B C 水模拟贪心