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 maxn = 1100000;
const int mod = 1e9 + 7;
int n, m;
char S[maxn];
int str[maxn];
int id[maxn], f[maxn], g[maxn];
struct pTree {
  int fail[maxn], trans[maxn][26], diff[maxn], nxt[maxn], len[maxn];
  int cnt, last;
  void init() {
    cnt = 1;
    last = 0;
    memset(trans[0], 0, sizeof trans[0]);
    memset(trans[1], 0, sizeof trans[1]);
    len[0] = 0, len[1] = -1;
    fail[0] = 1;
  }
  int newnode(int l) {
    ++cnt;
    memset(trans[cnt], 0, sizeof trans[cnt]);
    len[cnt] = l;
    return cnt;
  }
  int extend(int i) {
    int p = last, np, w = str[i];
    while (str[i - len[p] - 1] != str[i]) p = fail[p];
    if (!trans[p][w]) {
      np = newnode(len[p] + 2);
      int t = fail[p];
      while (str[i - len[t] - 1] != str[i]) t = fail[t];
      fail[np] = trans[t][w];
      diff[np] = len[np] - len[fail[np]];
      if (diff[np] == diff[fail[np]])
        nxt[np] = nxt[fail[np]];
      else
        nxt[np] = fail[np];
      trans[p][w] = np;
    } else
      np = trans[p][w];
    return last = np;
  }
} tr;
int main() {
  scanf("%s", S);
  n = strlen(S);
  if (n & 1) return puts("0"), 0;
  str[m = 0] = -1;
  for (int i = 0; i < n / 2; i++)
    str[++m] = S[i] - 'a', str[++m] = S[n - i - 1] - 'a';
  tr.init();
  for (int i = 1; i <= m; i++) id[i] = tr.extend(i);
  f[0] = 1;
  for (int i = 1; i <= m; i++) {
    for (int p = id[i]; p; p = tr.nxt[p]) {
      g[p] = f[i - tr.len[tr.nxt[p]] - tr.diff[p]];
      if (tr.diff[p] == tr.diff[tr.fail[p]]) (g[p] += g[tr.fail[p]]) %= mod;
      if (!(i & 1)) (f[i] += g[p]) %= mod;
    }
  }
  printf("%d\n", f[n]);
  return 0;
}