A. Diverse Substring

You are given a string ss, consisting of nn lowercase Latin letters.

A substring of string ss is a continuous segment of letters from ss. For example, "defor" is a substring of "codeforces" and "fors" is not.

The length of the substring is the number of letters in it.

Let's call some string of length nn diverse if and only if there is no letter to appear strictly more than n2n2 times. For example, strings "abc" and "iltlml" are diverse and strings "aab" and "zz" are not.

Your task is to find any diverse substring of string ss or report that there is none. Note that it is not required to maximize or minimize the length of the resulting substring.

Input

The first line contains a single integer nn (1≤n≤10001≤n≤1000) — the length of string ss.

The second line is the string ss, consisting of exactly nn lowercase Latin letters.

Output

Print "NO" if there is no diverse substring in the string ss.

Otherwise the first line should contain "YES". The second line should contain any diverse substring of string ss.

Examples

input

Copy

10
codeforces

output

Copy

YES
code

input

Copy

5
aaaaa

output

Copy

NO

題意：在字串s裡找一個子串s1，s1長度為L，其中s1裡任意一個元素出現次數不超過L/2次，問你能不能構造出這樣一個子串。

思路：推理一下規律可以發現，長串必然由短串構成，所以找最短的串即可，再推可以發現長度為2剛好滿足要求，每個字元只出現一次。

所以判斷相鄰元素是否有不同的，有則輸出，否則判定NO。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int n,flag;
char s[1010];
int main()
{
    while(~scanf("%d",&n))
    {
        scanf("%s",s);
        flag=0;
        for(int i=1;i<n;i++)
        {
            if(s[i-1]!=s[i])   //不同的相鄰元素
            {
                flag=1;
                printf("YES\n");
                printf("%c%c\n",s[i-1],s[i]);
                break;
            }
        }
        if(flag==0)printf("NO\n");
    }
    return 0;
}