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 MOD = 1000000007;
int g[1000010], f[1000010], trans[1000010][26];
int fail[1000010], len[1000010];
char st[1000010];
int c[1000010];
int d[1000010], pos[1000010];
int n, lst, cnt;
inline int mo(int x) {
  if (x >= MOD) return x - MOD;
  return x;
}
int getfail(int i, int x) {
  while (c[i - len[x] - 1] != c[i]) x = fail[x];
  return x;
}
int main() {
  scanf("%s", st + 1);
  n = strlen(st + 1);
  for (int i = 1; i <= n; i++)
    c[i] = (i & 1 ? st[(i + 1) >> 1] : st[n - (i >> 1) + 1]) - 'a';
  len[1] = -1;
  fail[0] = cnt = 1;
  f[0] = 1;
  c[0] = -1;
  for (int i = 1; i <= n; i++) {
    int x = getfail(i, lst);
    if (!trans[x][c[i]]) {
      int v = getfail(i, fail[x]);
      fail[++cnt] = trans[v][c[i]];
      trans[x][c[i]] = cnt;
      len[cnt] = len[x] + 2;
      d[cnt] = len[cnt] - len[fail[cnt]];
      pos[cnt] = d[cnt] == d[fail[cnt]] ? pos[fail[cnt]] : fail[cnt];
    }
    lst = trans[x][c[i]];
    for (x = lst; x; x = pos[x]) {
      g[x] = f[i - len[pos[x]] - d[x]];
      if (pos[x] != fail[x]) g[x] = mo(g[x] + g[fail[x]]);
      if (!(i & 1)) f[i] = mo(f[i] + g[x]);
    }
  }
  printf("%d\n", f[n]);
}