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 int read() {
  int data = 0, w = 1;
  char ch = getchar();
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') w = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
  return data * w;
}
constexpr int N(1e6 + 10), Mod(1e9 + 7);
inline void Add(int &x, const int &y) { x += y - Mod, x += x >> 31 & Mod; }
int n, son[N][26], fail[N], len[N], s[N], f[N], g[N];
int last = 0, cur = 1, cnt, dlt[N], pre[N];
char ch[N], a[N];
void Init() { len[1] = -1, fail[0] = 1, s[0] = 26; }
int find(int x) {
  while (s[cnt - len[x] - 1] != s[cnt]) x = fail[x];
  return x;
}
void extend(int c) {
  s[++cnt] = c;
  int t = find(last);
  if (!son[t][c]) {
    len[++cur] = len[t] + 2, fail[cur] = son[find(fail[t])][c], son[t][c] = cur,
    dlt[cur] = len[cur] - len[fail[cur]],
    pre[cur] = (dlt[cur] == dlt[fail[cur]] ? pre[fail[cur]] : fail[cur]);
  }
  last = son[t][c];
}
int main() {
  std::scanf("%s", ch + 1), n = std::strlen(ch + 1), f[0] = 1, Init();
  for (int i = 1, j = 0; i <= n && j <= n; i++)
    a[++j] = ch[i], a[++j] = ch[n - i + 1];
  for (int i = 1; i <= n; i++) {
    extend(a[i] - 'a');
    for (int x = last; x > 1; x = pre[x]) {
      g[x] = f[i - len[pre[x]] - dlt[x]];
      if (dlt[x] == dlt[fail[x]]) Add(g[x], g[fail[x]]);
      if (!(i & 1)) Add(f[i], g[x]);
    }
  }
  printf("%d\n", f[n]);
  return 0;
}