gym100971



Mihahim has a string s. He wants to delete exactly one character from it so that the resulting string would be a palindrome. Determine if he can do it, and if he can, what character should be deleted.

Input

The input contains a string s of length(2 ≤ |s| ≤ 200000), consisting of lowercase Latin letters.

Output

If the solution exists, output «YES» (without quotes) in the first line. Then in the second line output a single integerx — the number of the character that should be removed froms so that the resulting string would be a palindrome. The characters in the string are numbered from 1. If there are several possible solutions, output any of them.

If the solution doesn't exist, output «NO» (without quotes).

Example

Input

evertree

Output

YES2

Input

emerald

Output

NO

Input

aa

Output

YES2

给你一个串然后问你删除一个字符能不能形成回文串

ac代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;char a[1000005];int main(){    while(gets(a))    {       int len=strlen(a);       int sum=0;       int p=0,q=len-1;       int k;       while(p<=q)       {           if(a[p]==a[q])           {               p++;q--;           }           else           {               sum++;               int sum1=0,sum2=0;               int r=p,t=q-1;               while(r<=t)               {                   if(a[r]==a[t])                   {                       r++;t--;                   }                   else                   {                       sum1++;break;                   }               }               r=p+1;t=q;               while(r<=t)               {                  if(a[r]==a[t])                   {                       r++;t--;                   }                   else                   {                       sum2++;break;                   }               }               sum+=min(sum1,sum2);               if(sum1>sum2)               k=p;               else               k=q;               break;           }       }       if(sum==0)       {           printf("YES\n");           printf("%d\n",len/2+1);       }       if(sum==1)       {           printf("YES\n");           printf("%d\n",k+1);       }       if(sum==2)       printf("NO\n");    }    return 0;}