K. Palindromization

time limit per test

2.0 s

memory limit per test

256 MB

input

standard input

output

standard output

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 integer x — the number of the character that should be removed from s 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).

Examples

input

evertree

output

YES 2

input

emerald

output

NO

input

aa

output

YES 2

 题意：给你一个字符串，删除一个字符，是否形成回文，如果有，输出删除的位置；

思路：对于删除一个字符以后，这个字符串的对称轴的位置只有两个，所有枚举对称轴，复杂度O(n);

#include
             
              #include
              
               #include
               
                #include
                
                 #include
                 
                  #include
                  
                   using namespace std;#define LL long longconst int N=2e5+100,M=1e6+10,inf=1e9+10;const LL INF=1e18+10,mod=19260817;char a[N];int checkl(int l,int r,int n,int L){ while(r<=n) { while(a[l]!=a[r]) { if(L!=1)return 0; L=l; l--; } l--;r++; } return L;}int checkr(int l,int r,int n,int R){ while(l>=1) { while(a[l]!=a[r]) { if(R!=n)return 0; else R=r; r++; } l--;r++; } return R;}int main(){ scanf("%s",a+1); int n=strlen(a+1); if(n%2==0) { int ans=checkl(n/2,n/2+2,n,1); if(ans)return 0*printf("YES\n%d\n",ans); ans=checkr(n/2-1,n/2+1,n,n); if(ans)return 0*printf("YES\n%d\n",ans); puts("NO"); } else { int ans=checkl(n/2+1,n/2+2,n,1); if(ans)return 0*printf("YES\n%d\n",ans); ans=checkr(n/2,n/2+1,n,n); if(ans)return 0*printf("YES\n%d\n",ans); puts("NO\n"); } return 0;}