Palindrome Degree（hash的思想题）

个人心得：这题就是要确定是否为回文串，朴素算法会超时，所以想到用哈希，哈希从左到右和从右到左的key值一样就一定是回文串，

那么问题来了，正向还能保证一遍遍历，逆向呢，卡住我了，后面发现网上大神的秦九韶算法用上了，厉害了。

关于哈希，这题用的是最简单的哈希思想，就是用M取合理的素数，这题取得是3，

然后正向就是 a+=a*M+ch[i]。逆向用秦九韶可以在循环的时候算出来，

举个例子比如三个值1，2，3。

假如逆向开始 则 (((0+s[3])*M+s[2])*M+s[1]);化简就等于s[1]+s[2]*M+s[3]*M^2;

哇，规律就来了，那么在正向就能确定逆向的值，服气服气

题目:

String s of length n is called k-palindrome, if it is a palindrome itself, and its prefix and suffix of length  are (k - 1)-palindromes. By definition, any string (even empty) is 0-palindrome.

Let's call the palindrome degree of string s such a maximum number k, for which s is k-palindrome. For example, "abaaba" has degree equals to 3.

You are given a string. Your task is to find the sum of the palindrome degrees of all its prefixes.

Input

The first line of the input data contains a non-empty string, consisting of Latin letters and digits. The length of the string does not exceed 5·10⁶. The string is case-sensitive.

Output

Output the only number — the sum of the polindrome degrees of all the string's prefixes.

Example

Input

a2A

Output

1

Input

abacaba

Output

6

#include<iostream>
#include<cstdio>
#include<cmath>
#include<set>
#include<vector>
#include<cstring>
#include<iomanip>
#include<algorithm>
using namespace std;
#define inf 1<<29
#define nu 4000005
#define maxnum 5000005
#define num 2005
int n;
const long long N = 13, M = 3;
char ch[maxnum];
long long number[maxnum];
long long sum;
int main()
{
    scanf("%s",ch);
    int len=strlen(ch);
    long long a=0,b=0,p=1;
    memset(number,0,sizeof(number));
    for(int i=0;i<len;i++){
        a=a*M+ch[i];b+=ch[i]*p;p*=M;
        if(a==b) sum+=number[i+1]=(number[(i+1)/2]+1);
    }
    printf("%lld\n",sum);
    return 0;
}