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;
inline int read() {
  int X = 0, w = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    X = (X << 3) + (X << 1) + ch - '0', ch = getchar();
  return X * w;
}
char ch[1000007], s[1000007];
int n, last, tot, anc[1000007], diz[1000007], ans[1000007], f[1000007];
struct mmp {
  int z, len, son[26];
} t[1000007];
void extend(int c, int n, char *s) {
  int p = last;
  while (s[n - t[p].len - 1] != s[n]) {
    p = t[p].z;
  }
  if (!t[p].son[c]) {
    int v = ++tot, k = t[p].z;
    t[v].len = t[p].len + 2;
    while (s[n - t[k].len - 1] != s[n]) {
      k = t[k].z;
    }
    t[v].z = t[k].son[c];
    t[p].son[c] = v;
    diz[v] = t[v].len - t[t[v].z].len;
    anc[v] = (diz[v] == diz[t[v].z]) ? anc[t[v].z] : t[v].z;
  }
  last = t[p].son[c];
}
int main() {
  t[tot = 1].len = -1;
  t[0].z = t[1].z = 1;
  anc[0] = 1;
  scanf("%s", ch + 1);
  n = strlen(ch + 1);
  if (n & 1) {
    puts("0");
    return 0;
  }
  for (int i = 1; i <= n; i += 2) {
    s[i] = ch[(i + 1) / 2];
  }
  reverse(&ch[1], &ch[n + 1]);
  for (int i = 2; i <= n; i += 2) {
    s[i] = ch[(i + 1) / 2];
  }
  ans[0] = 1;
  for (int i = 1; i <= n; i++) {
    extend(s[i] - 97, i, s);
    for (int k = last; k; k = anc[k]) {
      f[k] = ans[i - t[anc[k]].len - diz[k]];
      if (anc[k] != t[k].z) {
        f[k] = (f[k] + f[t[k].z]) % 1000000007;
      }
      if (!(i & 1)) {
        ans[i] = (ans[i] + f[k]) % 1000000007;
      }
    }
  }
  printf("%d\n", ans[n]);
  return 0;
}