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>
inline long long gi() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) f ^= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f ? x : -x;
}
char _s[1000010], s[1000010];
int n;
int ch[1000010][26], dif[1000010], anc[1000010], fail[1000010], len[1000010],
    cnt, f[1000010], g[1000010], lst;
inline int newnode(int Len) {
  ++cnt, len[cnt] = Len;
  return cnt;
}
inline void extend(int c, int n) {
  int p = lst, q = ch[p][c];
  while (s[n - len[p] - 1] != s[n]) p = fail[p];
  if (!ch[p][c]) {
    int np = newnode(len[p] + 2);
    q = fail[p];
    while (s[n - len[q] - 1] != s[n]) q = fail[q];
    fail[np] = ch[q][c];
    ch[p][c] = np;
    dif[np] = len[np] - len[fail[np]];
    anc[np] = dif[np] == dif[fail[np]] ? anc[fail[np]] : fail[np];
  }
  lst = ch[p][c];
}
int main() {
  scanf("%s", _s + 1);
  n = strlen(_s + 1);
  for (int l = 1, r = n, p = 1; l <= r; ++l, --r)
    s[p++] = _s[l], s[p++] = _s[r];
  cnt = lst = 1;
  len[0] = 0, len[1] = -1;
  fail[0] = fail[1] = 1;
  f[0] = 1;
  for (int i = 1; i <= n; ++i) {
    extend(s[i] - 'a', i);
    for (int j = lst; j; j = anc[j]) {
      g[j] = f[i - len[anc[j]] - dif[j]];
      if (anc[j] ^ fail[j]) g[j] = (g[j] + g[fail[j]]) % 1000000007;
      if (~i & 1) f[i] = (f[i] + g[j]) % 1000000007;
    }
  }
  printf("%d\n", f[n]);
  return 0;
}