Codeforces Round #358 (Div. 2) A B C 水 水 dfs序+dp

A. Alyona and Numbers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.

Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and equals 0.

As usual, Alyona has some troubles and asks you to help.

Input

The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).

Output

Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.

Examples

Input

6 12

Output

14

Input

11 14

Output

31

Input

1 5

Output

1

Input

3 8

Output

5

Input

5 7

Output

7

Input

21 21

Output

88

Note

Following pairs are suitable in the first sample case:

for x = 1 fits y equal to 4 or 9;

• for x = 2 fits y equal to 3 or 8;
• for x = 3 fits y equal to 2, 7 or 12;
• for x = 4 fits y equal to 1, 6 or 11;
• for x = 5 fits y equal to 5 or 10;
• for x = 6 fits y equal to 4 or 9.

Only the pair (1, 4) is suitable in the third sample case.

题意：给你 n，m     1 ≤ x ≤ n, 1 ≤ y ≤ m  问有多少对x/y 使得（x+y）%5==0

题解：水

#include <bits/stdc++.h>
#define ll  __int64
using namespace  std;
int n,m;
int main()
{
    scanf("%d %d",&n,&m);
    ll ans=0;
    for(int i=1;i<=n;i++)
    {
        int exm=i%5;
        int re=5-exm;
        if(m>=re){
        ans=ans+(m-re)/5;
        ans++;
        }
    }
    cout<<ans<<endl;
    return 0;
}

B. Alyona and Mex

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Someone gave Alyona an array containing n positive integers a₁, a₂, ..., a_n. In one operation, Alyona can choose any element of the array and decrease it, i.e. replace with any positive integer that is smaller than the current one. Alyona can repeat this operation as many times as she wants. In particular, she may not apply any operation to the array at all.

Formally, after applying some operations Alyona will get an array of n positive integers b₁, b₂, ..., b_n such that 1 ≤ b_i ≤ a_i for every 1 ≤ i ≤ n. Your task is to determine the maximum possible value of mex of this array.

Mex of an array in this problem is the minimum positive integer that doesn't appear in this array. For example, mex of the array containing 1, 3 and 4 is equal to 2, while mex of the array containing 2, 3 and 2 is equal to 1.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of elements in the Alyona's array.

The second line of the input contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — the elements of the array.

Output

Print one positive integer — the maximum possible value of mex of the array after Alyona applies some (possibly none) operations.

Examples

Input

5
1 3 3 3 6

Output

5

Input

2
2 1

Output

3

Note

In the first sample case if one will decrease the second element value to 2 and the fifth element value to 4 then the mex value of resulting array 1 2 3 3 4 will be equal to 5.

To reach the answer to the second sample case one must not decrease any of the array elements.

题意：给你n个数 每次操作只能减小某个数 经过一系列操作之后 使得没有出现的数的最小值尽量大

题解：水

#include <bits/stdc++.h>
#define ll  __int64
using namespace  std;
int n;
int a[100005];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    sort(a+1,a+1+n);
    int now=1;
    for(int i=1;i<=n;i++)
    {
        if(now<=a[i])
            now++;
        else
            continue;
    }
    cout<<now<<endl;
    return 0;
}

C. Alyona and the Tree

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Alyona decided to go on a diet and went to the forest to get some apples. There she unexpectedly found a magic rooted tree with root in the vertex 1, every vertex and every edge of which has a number written on.

The girl noticed that some of the tree's vertices are sad, so she decided to play with them. Let's call vertex v sad if there is a vertex u in subtree of vertex v such that dist(v, u) > a_u, where a_u is the number written on vertex u, dist(v, u) is the sum of the numbers written on the edges on the path from v to u.

Leaves of a tree are vertices connected to a single vertex by a single edge, but the root of a tree is a leaf if and only if the tree consists of a single vertex — root.

Thus Alyona decided to remove some of tree leaves until there will be no any sad vertex left in the tree. What is the minimum number of leaves Alyona needs to remove?

Input

In the first line of the input integer n (1 ≤ n ≤ 10⁵) is given — the number of vertices in the tree.

In the second line the sequence of n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) is given, where a_i is the number written on vertex i.

The next n - 1 lines describe tree edges: i^th of them consists of two integers p_i and c_i (1 ≤ p_i ≤ n,  - 10⁹ ≤ c_i ≤ 10⁹), meaning that there is an edge connecting vertices i + 1 and p_i with number c_i written on it.

Output

Print the only integer — the minimum number of leaves Alyona needs to remove such that there will be no any sad vertex left in the tree.

Example

Input

9
88 22 83 14 95 91 98 53 11
3 24
7 -8
1 67
1 64
9 65
5 12
6 -80
3 8

Output

5

Note

The following image represents possible process of removing leaves from the tree:

                

题意：给你一个树 有点权 边权 删除最少的叶子结点 使得 对于任意一个点到其子树中的点的路径边权和小于等与目标点的点权

题解：dfs处理 记录当前节点的祖先节点中到当前节点的路径边权和的最大值

#include <bits/stdc++.h>
#define ll  __int64
using namespace  std;
ll n;
ll a[100005];
ll p[100005];
ll in[100005];
ll out[100005];
ll d[100005];
map<ll,ll> mp;
ll nedge=0;
ll dfn=0;
ll re=0;
struct node
{
    ll to;
    ll pre;
    ll we;
}N[200005];
void add(ll pre,ll to,ll we)
{
    nedge++;
    N[nedge].to=to;
    N[nedge].we=we;
    N[nedge].pre=p[pre];
    p[pre]=nedge;
}
void getdfs(ll root,ll sum,ll now)
{
    in[root]=++dfn;
    d[root]=max(sum,now);
    sum=d[root];
    mp[root]=1;
    for(int i=p[root];i;i=N[i].pre){
        if(mp[N[i].to])
            continue;
        getdfs(N[i].to,sum+N[i].we,N[i].we);
    }
    out[root]=dfn;
}
void dfs(ll root)
{
    mp[root]=1;
    for(int i=p[root];i;i=N[i].pre){
        if(mp[N[i].to])
            continue;
        if(d[N[i].to]>a[N[i].to])
        {
            re=re+out[N[i].to]-in[N[i].to]+1;
            continue ;
        }
        dfs(N[i].to);
    }
}
int main()
{
    ll r,w;
    memset(p,0,sizeof(p));
    scanf("%I64d",&n);
    for(ll i=1;i<=n;i++)
        scanf("%I64d",&a[i]);
    for(ll i=2;i<=n;i++)
    {
        scanf("%I64d %I64d",&r,&w);
        add(i,r,w);
        add(r,i,w);
    }
    getdfs(1,0ll,0ll);
    mp.clear();
    dfs(1);
    printf("%I64d
",re);
    return 0;
}