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;
namespace zzc {
const int mod = 1e9 + 7;
const int maxn = 1e6 + 5;
int len[maxn], fail[maxn], g[maxn], f[maxn], slink[maxn], dif[maxn],
    trans[maxn][26];
int n, tot, lst, le;
char ch[maxn], s[maxn];
inline void init() {
  len[0] = 0;
  len[1] = -1;
  fail[0] = 1;
  fail[1] = 0;
  tot = 1;
  lst = 0;
}
inline int get_fail(int x) {
  while (ch[le - len[x] - 1] != ch[le]) x = fail[x];
  return x;
}
inline void insert(char c) {
  ch[++le] = c;
  int p = get_fail(lst);
  if (!trans[p][c - 'a']) {
    int cur = ++tot;
    len[cur] = len[p] + 2;
    int tmp = get_fail(fail[p]);
    fail[cur] = trans[tmp][c - 'a'];
    trans[p][c - 'a'] = cur;
    dif[cur] = len[cur] - len[fail[cur]];
    if (dif[cur] == dif[fail[cur]])
      slink[cur] = slink[fail[cur]];
    else
      slink[cur] = fail[cur];
  }
  lst = trans[p][c - 'a'];
}
void work() {
  init();
  scanf("%s", ch + 1);
  n = strlen(ch + 1);
  for (register int i = 1, j = 0; i <= (n + 1) >> 1; i++)
    s[++j] = ch[i], s[++j] = ch[n - i + 1];
  f[0] = 1;
  for (register int i = 1; i <= n; i++) {
    insert(s[i]);
    for (register int x = lst; x > 1; x = slink[x]) {
      g[x] = f[i - len[slink[x]] - dif[x]];
      if (dif[x] == dif[fail[x]]) g[x] = (g[x] + g[fail[x]]) % mod;
      if (i % 2 == 0) f[i] = (f[i] + g[x]) % mod;
    }
  }
  printf("%d\n", f[n]);
}
}  // namespace zzc
int main() {
  zzc::work();
  return 0;
}