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;
int n;
char s[1000006], origin[1000006];
int ch[1000006][27], fail[1000006], diff[1000006], anc[1000006], len[1000006],
    total, last;
int f[1000006], g[1000006];
const int mod = 1e9 + 7;
void init() {
  s[0] = '#';
  len[1] = -1;
  fail[0] = total = 1;
}
int getfail(int u, int n) {
  while (s[n - len[u] - 1] != s[n]) u = fail[u];
  return u;
}
void append(int c, int n) {
  int v = getfail(last, n);
  if (!ch[v][c]) {
    int u = ++total;
    len[u] = len[v] + 2;
    fail[u] = ch[getfail(fail[v], n)][c];
    ch[v][c] = u;
    diff[u] = len[u] - len[fail[u]];
    anc[u] = diff[u] == diff[fail[u]] ? anc[fail[u]] : fail[u];
  }
  last = ch[v][c];
}
int main() {
  scanf("%s", origin + 1);
  init();
  n = (int)strlen(origin + 1);
  if (n & 1) {
    puts("");
    return 0;
  }
  for (int i = 1; i <= n; i += 2) s[i] = origin[i / 2 + 1];
  for (int i = 2; i <= n; i += 2) s[i] = origin[n - i / 2 + 1];
  g[0] = 1;
  for (int i = 1; i <= n; ++i) {
    append(s[i] - 'a' + 1, i);
    for (int j = last; j; j = anc[j]) {
      f[j] = g[i - diff[j] - len[anc[j]]];
      if (anc[j] != fail[j]) f[j] = (f[j] + f[fail[j]]) % mod;
      if (i & 1) continue;
      g[i] = (g[i] + f[j]) % mod;
    }
  }
  printf("%d\n", g[n]);
}