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 = , M = ;
 char ch[maxnum];
 long long number[maxnum];
 long long sum;
 int main()
 {
     scanf("%s",ch);
     int len=strlen(ch);
     long long a=,b=,p=;
     memset(number,,sizeof(number));
     for(int i=;i<len;i++){
         a=a*M+ch[i];b+=ch[i]*p;p*=M;
         if(a==b) sum+=number[i+]=(number[(i+)/]+);
     }
     printf("%lld\n",sum);
     return ;
 }