Codeforces Round #505 (rated, Div. 1 + Div. 2, based on VK Cup 2018 Final)

　　第一次点进去，结果忘记参赛了。。。但还是做了2个题（水平低到没底。。请轻虐。。）下面是题和我做的不确定是否正确的解决方案。

　　原题链接：http://codeforces.com/contest/1025

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A. Doggo Recoloring

time limit per test:

1 second

memory limit per test:

256 megabytes

input:

standard input

output:

standard output

Panic is rising in the committee for doggo standardization — the puppies of the new brood have been born multi-colored! In total there are 26 possible colors of puppies in the nature and they are denoted by letters from 'a' to 'z' inclusive.

The committee rules strictly prohibit even the smallest diversity between doggos and hence all the puppies should be of the same color. Thus Slava, the committee employee, has been assigned the task to recolor some puppies into other colors in order to eliminate the difference and make all the puppies have one common color.

Unfortunately, due to bureaucratic reasons and restricted budget, there's only one operation Slava can perform: he can choose a color $\mathrm{xxsuch\; that\; there\; are\; currently at\; least\; two puppies\; of\; color xx and\; recolor all puppies\; of\; the\; color xx into\; some\; arbitrary\; color yy.\; Luckily,\; this\; operation\; can\; be\; applied\; multiple\; times\; (including\; zero).}$

For example, if the number of puppies is $\mathrm{77 and\; their\; colors\; are\; represented\; as\; the\; string\; "abababc",\; then\; in\; one\; operation\; Slava\; can\; get\; the\; results\; "zbzbzbc",\; "bbbbbbc",\; "aaaaaac",\; "acacacc"\; and\; others.\; However,\; if\; the\; current\; color\; sequence\; is\; "abababc",\; then\; he\; can\text{'}t\; choose xx=\text{'}c\text{'}\; right\; now,\; because\; currently\; only\; one\; puppy\; has\; the\; color\; \text{'}c\text{'}.}$

Help Slava and the committee determine whether it is possible to standardize all the puppies, i.e. after Slava's operations all the puppies should have the same color.

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 1051\le n\le 105) —\; the\; number\; of\; puppies.}$

The second line contains a string $\mathrm{ss of\; length nn consisting\; of\; lowercase\; Latin\; letters,\; where\; the ii-th\; symbol\; denotes\; the ii-th\; puppy\text{'}s\; color.}$

Output

If it's possible to recolor all puppies into one color, print "Yes".

Otherwise print "No".

Output the answer without quotation signs.

Examples

input

6
aabddc

output

Yes

input

3
abc

output

No

input

3
jjj

output

Yes

Note

In the first example Slava can perform the following steps:

1. take all puppies of color 'a' (a total of two) and recolor them into 'b';
2. take all puppies of color 'd' (a total of two) and recolor them into 'c';
3. take all puppies of color 'b' (three puppies for now) and recolor them into 'c'.

In the second example it's impossible to recolor any of the puppies.

In the third example all the puppies' colors are the same; thus there's no need to recolor anything.

#include <bits/stdc++.h>
#define Index(ch) (ch - 'a')
const int N = 100000;
int has[26];
void init(char*c, int s)
{
    memset(has, 0, 26);
    for (int i = 0; i < s; i++)
        has[Index(c[i])]++;
}
int main()
{
    char cors[N];int dogs;
    while (~scanf("%d", &dogs))
    {
        if (dogs > 1 && dogs < N)
        {
            scanf("%s", cors);
            init(cors, dogs);
            for (int i = 0; i < 26; i++)
                if (has[i] >= 2)
                {
                    dogs = 0;
                    break;
                }
            if (!dogs)
                puts("Yes");
            else
                puts("No");
        }
        else
            puts("No");
    }
    return 0;
}

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B. Weakened Common Divisor

time limit per test:

1.5 seconds

memory limit per test:

256 megabytes

input:

standard input

output:

standard output

During the research on properties of the greatest common divisor (GCD) of a set of numbers, Ildar, a famous mathematician, introduced a brand new concept of the weakened common divisor (WCD) of a list of pairs of integers.

For a given list of pairs of integers $(a1,b1)(a1,b1), (a2,b2)(a2,b2),\; ..., (an,bn)(an,bn) their WCD is\; arbitrary\; integer\; greater\; than 11,\; such\; that\; it\; divides\; at\; least\; one\; element\; in\; each\; pair.\; WCD\; may\; not\; exist\; for\; some\; lists.$

For example, if the list looks like $[(12,15),(25,18),(10,24)][(12,15),(25,18),(10,24)],\; then\; their\; WCD\; can\; be\; equal\; to 22, 33, 55 or 66 (each\; of\; these\; numbers\; is\; strictly\; greater\; than 11 and\; divides\; at\; least\; one\; number\; in\; each\; pair).$

You're currently pursuing your PhD degree under Ildar's mentorship, and that's why this problem was delegated to you. Your task is to calculate WCD efficiently.

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 1500001\le n\le 150000) —\; the\; number\; of\; pairs.}$

Each of the next $\mathrm{nn lines\; contains\; two\; integer\; values aiai, bibi (2\le ai,bi\le 2\cdot 1092\le ai,bi\le 2\cdot 109).}$

Output

Print a single integer — the WCD of the set of pairs.

If there are multiple possible answers, output any; if there is no answer, print $-1-1.$

Examples

input

3
17 18
15 24
12 15

output

6

input

2
10 16
7 17

output

-1

input

5
90 108
45 105
75 40
165 175
33 30

output

5

Note

In the first example the answer is $\mathrm{66 since\; it\; divides 1818 from\; the\; first\; pair, 2424 from\; the\; second\; and 1212 from\; the\; third\; ones.\; Note\; that\; other\; valid\; answers\; will\; also\; be\; accepted.}$

In the second example there are no integers greater than $\mathrm{11 satisfying\; the\; conditions.}$

In the third example one of the possible answers is $\mathrm{55.\; Note\; that,\; for\; example, 1515 is\; also\; allowed,\; but\; it\text{'}s\; not\; necessary\; to\; maximize\; the\; output.}$

#include <bits/stdc++.h>
const int N = 150000;
int a[N], b[N];
int gcd(int a, int b)
{
    if (b == 0) return a;
    return gcd(b, a%b);
}
int main()
{
    int s;
    while (~scanf("%d", &s))
    {
        int p1, p2, p3, p4, f1 = 0, f2 = 0, f3 = 0, f4 = 0;
        for (int i = 0; i < s; i++)
            scanf("%d %d", &a[i], &b[i]);
        p1 = gcd(a[0], a[1]);
        p2 = gcd(a[0], b[1]);
        p3 = gcd(b[0], a[1]);
        p4 = gcd(b[0], b[1]);

        if (p1 > 1)
        {
            f1 = p1;
            for (int i = 2; i < s; i++)
                p1 = gcd(p1, a[i]);
            for (int i = 2; i < s; i++)
                f1 = gcd(f1, b[i]);
        }
        if (p2 > 1)
        {
            f2 = p2;
            for (int i = 2; i < s; i++)
                p2 = gcd(p2, a[i]);
            for (int i = 2; i < s; i++)
                f2 = gcd(f2, b[i]);
        }
        if (p3 > 1)
        {
            f3 = p3;
            for (int i = 2; i < s; i++)
                p3 = gcd(p3, a[i]);
            for (int i = 2; i < s; i++)
                f3 = gcd(f3, b[i]);
        }
        if (p4 > 1)
        {
            f4 = p4;
            for (int i = 2; i < s; i++)
                p4 = gcd(p4, a[i]);
            for (int i = 2; i < s; i++)
                f4 = gcd(f4, b[i]);
        }
        if (p1 > 1) printf("%d
", p1);
        else if (f1 > 1) printf("%d
", f1);
        else if (p2 > 1) printf("%d
", p2);
        else if (f2 > 1) printf("%d
", f2);
        else if (p3 > 1) printf("%d
", p3);
        else if (f3 > 1) printf("%d
", f3);
        else if (p4 > 1) printf("%d
", p4);
        else if (f4 > 1) printf("%d
", f4);
        else
            puts("-1");
    }
    return 0;
}

---

D. Recovering BST

time limit per test:

1 second

memory limit per test:

256 megabytes

input:

standard input

output:

standard output

Dima the hamster enjoys nibbling different things: cages, sticks, bad problemsetters and even trees!

Recently he found a binary search tree and instinctively nibbled all of its edges, hence messing up the vertices. Dima knows that if Andrew, who has been thoroughly assembling the tree for a long time, comes home and sees his creation demolished, he'll get extremely upset.

To not let that happen, Dima has to recover the binary search tree. Luckily, he noticed that any two vertices connected by a direct edge had their greatest common divisor value exceed $11.$

Help Dima construct such a binary search tree or determine that it's impossible. The definition and properties of a binary search tree can be found here.

Input

The first line contains the number of vertices $\mathrm{nn (2\le n\le 7002\le n\le 700).}$

The second line features $\mathrm{nn distinct\; integers aiai (2\le ai\le 1092\le ai\le 109) —\; the\; values\; of\; vertices in\; ascending\; order.}$

Output

If it is possible to reassemble the binary search tree, such that the greatest common divisor of any two vertices connected by the edge is greater than $\mathrm{11,\; print\; "Yes"\; (quotes\; for\; clarity).}$

Otherwise, print "No" (quotes for clarity).

Examples

input

6
3 6 9 18 36 108

output

Yes

input

2
7 17

output

No

input

9
4 8 10 12 15 18 33 44 81

output

Yes

Note

The picture below illustrates one of the possible trees for the first example.

The picture below illustrates one of the possible trees for the third example.

（还在推dp方程中。。。）方程是推不出来了（> <），直接看别人的题解吧：

　　http://codeforces.com/contest/1025/submission/41898155 Orz...

　　这里还有一份是循环的区间dp的解法 Orz....：

#include<bits/stdc++.h>
int n,s[704],g[705][705],L[705][705],R[705][705],f[705][705];
int main()
{
    for(int i = 0;i <= (n=s[0]); i++)
        scanf("%d",&s[i]);

    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            g[i][j]=std::__gcd(s[i],s[j])>1;

    for(int i=1;i<=n;i++)
        L[i][i]=R[i][i]=1;

    for(int i=n;i>=1;i--)
        for(int j=i;j<=n;j++)
            for(int k=i;k<=j;k++)
                if(L[i][k] && R[k][j])
                    f[i][j] |= 1,
                    R[i-1][j] |= g[k][i-1],
                    L[i][j+1] |= g[k][j+1];

    return puts(f[1][n]?"Yes":"No"),0;
}