Gym_100971k_Palindromization （回文串）

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).

Example

Input

evertree

Output

YES2

Input

emerald

Output

NO

Input

aa

Output

YES2

题意：

给你一个串，问你在这个串中删除一个字符后这个串能否成为一个回文串，如果能还得输出需要删除的字符的位置。

解：

从左往右找，碰到第一个左右不同的字符时记一下位置，然后重新传一个删掉左或右字符后的字符串进行Manacher一下就行。

需要特判的是如果这个串本身就是回文串的处理。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>using namespace std;const int maxn=200005;char str0[maxn],str[maxn<<1],str1[maxn];int p[maxn<<1],len;int l,r;void start(){    str[0]='$',str[1]='#';    for(int i=0;i<len;i++)    {        str[i*2+2]=str0[i];        str[i*2+3]='#';    }    str[len*2+2]='\0';}int solve(){    memset(p,0,sizeof(p));    int mx=0,id,ans=1;    for(int i=1;i<len*2+2;++i)    {        if(mx>i)p[i]=min(p[2*id-i],mx-i);        else p[i]=1;        for(;str[i-p[i]]==str[i+p[i]];p[i]++);        if(p[i]+i>mx)mx=p[i]+i,id=i;        if(p[i]>ans)ans=p[i];    }    return --ans;}int main(){    int flag=0;    while(~scanf("%s",str1))    {        flag=0;int k;        int len1=strlen(str1);        int j=0;        for(int i=0;i<len1;i++)        {            str0[j]=str1[i];            j++;        }        len=j;        start();        if(solve()==len1)        {flag=1;k=len1/2+1;}        else{        for(int i=0;i<len1/2+1;i++)        {            if(str1[i]!=str1[len1-1-i])            {                l=i;r=len1-1-i;break;            }        }        int j=0;        for(int i=0;i<len1;i++)        {            if(i!=l)            {str0[j]=str1[i];j++;}        }        len=j;        start();        int ans=solve();        if(ans==len){flag=1;k=l+1;}            j=0;            for(int i=0;i<len1;i++)            {                if(i!=r)                {str0[j]=str1[i];j++;}            }            len=j;            start();            ans=solve();            if(ans==len){flag=1;k=r+1;}        }        //printf("L:%d R:%d K:%d",l,r,k);        if(flag==1)printf("YES\n%d\n",k);        else        printf("NO\n");    }    return 0;}

<-------!    额，这道题也可以不用Manacher，直接跑一边就行，但比赛时没想着么多，直接就用Manacher感觉写起来简单点   !------->