codeforces 7D

D. Palindrome Degree

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

Input

a2A

Output

1

Input

abacaba

Output

6

（一道好题，我想到的是用manacher做，听说还可以用kmp和hash做

题意：遍历给定字符串的所有前缀，如果该前缀是个回文串，将该串对半分看还是不是个回文串，如果是，则其degree+1，如cc是个回文串，c也是个回文串，所以cc的degree==1

解题思路:跑一边manacher，用p数组判断前缀是不是回文串，如果是回文串，则它的degree是其对半分后串的degree+1。

ac代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <string>
 4 #include <algorithm>
 5 using namespace std;
 6 typedef long long ll;
 7 const int maxn = 5*1e6+10;
 8 char st[maxn];
 9 char s[maxn<<1];
10 int p[maxn<<1];
11 int ans[maxn];
12 void init(int le) {
13     for(int i=le-1;i>=0;--i) {
14         s[2*i+2]=st[i];
15         s[2*i+3]='*';
16     }
17     s[0]='$';
18     s[1]='*';
19 }
20 void manacher(int le) {
21     int mx=0,id=0;
22     for(int i=1;i<2*le+2;++i) {
23         p[i]=mx>i?min(p[id*2-i],mx-i):1;
24         while(s[i+p[i]]==s[i-p[i]]) p[i]++;
25         if(i+p[i]>mx) {
26             mx=i+p[i];
27             id=i;
28         }
29     }
30 }
31 int main() {
32      ll res=0;
33      scanf("%s",st);
34      int le = strlen(st);
35      init(le);
36      manacher(le);
37      int u=2;
38      for(int i=0;i<le;++i) {
39          if(p[i+2]!=i+2) {
40              ans[i]=0;
41          }
42          else {
43              ans[i]=ans[(i-1)/2]+1;
44              ans[i]=max(ans[i],1);
45          }
46          res+=ans[i];
47      }
48      printf("%lld\n",res);
49 }
50
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53