Problem: Given a string s, find the number of ways to split s to substrings such that if there are k substrings (p1, p2, p3, ..., pk) in partition, then pi = pk - i + 1 for all i (1 ≤ i ≤ k) and k is even.

Since the number of ways can be large, print it modulo 109 + 7.

Input

The only line of input contains a string s (2 ≤ |s| ≤ 106) of even length consisting of lowercase Latin letters. 

Output

Print one integer, the number of ways of partitioning the string modulo 109 + 7.

Examples

Input

abcdcdab


Output

1

Input

abbababababbab


Output

3

Note

In the first case, the only way to partition the string is ab|cd|cd|ab.

In the second case, the string can be partitioned as ab|b|ab|ab|ab|ab|b|ab or ab|b|abab|abab|b|ab or abbab|ab|ab|abbab.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 50;
const int mod = 1e9 + 7;
int n, f[N], tot = 1;
char s[N], t[N];
int last, son[N][27], diff[N], len[N], link[N], slink[N], series_ans[N];
inline int getfail(int pos, int l) {
  while (s[l - len[pos] - 1] != s[l]) pos = link[pos];
  return pos;
}
int main() {
  len[1] = -1;
  link[0] = 1;
  diff[0] = 1;
  scanf("%s", t + 1);
  n = strlen(t + 1);
  if (n & 1) {
    puts("0");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (i & 1)
      s[i] = t[(i + 1) / 2];
    else
      s[i] = t[n - i / 2 + 1];
  }
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    int c = s[i] - 'a' + 1;
    int now = getfail(last, i);
    if (!son[now][c]) {
      len[++tot] = len[now] + 2;
      link[tot] = son[getfail(link[now], i)][c];
      diff[tot] = len[tot] - len[link[tot]];
      if (diff[tot] == diff[link[tot]])
        slink[tot] = slink[link[tot]];
      else
        slink[tot] = link[tot];
      son[now][c] = tot;
    }
    last = son[now][c];
    for (int v = last; len[v]; v = slink[v]) {
      series_ans[v] = f[i - (len[slink[v]] + diff[v])];
      if (diff[v] == diff[link[v]])
        series_ans[v] = (series_ans[v] + series_ans[link[v]]) % mod;
      f[i] = (f[i] + series_ans[v]) % mod;
    }
    if (i & 1) f[i] = 0;
  }
  printf("%d\n", f[n]);
}