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 Imx = 2147483647;
const long long Lbig = 2e18;
const int mod = 1e9 + 7;
const int maxn = 1000111;
char s[maxn];
int n, tot, go[maxn][26], fail[maxn], slink[maxn], len[maxn], dlen[maxn];
int st[maxn];
int append(int p, int l) {
  int c = s[l] - 'a';
  while (s[l - len[p] - 1] != s[l]) p = fail[p];
  if (!go[p][c]) {
    int np = ++tot;
    len[np] = len[p] + 2;
    int t = fail[p];
    while (s[l - len[t] - 1] != s[l]) t = fail[t];
    fail[np] = go[t][c];
    dlen[np] = len[np] - len[fail[np]];
    slink[np] = dlen[np] == dlen[fail[np]] ? slink[fail[np]] : fail[np];
    go[p][c] = np;
  }
  return go[p][c];
}
int dp[maxn], gdp[maxn];
void upd(int &x, int v) { x = x + v >= mod ? x + v - mod : x + v; }
int main() {
  static char ss[maxn];
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = 1; i <= n / 2; i++) ss[i * 2 - 1] = s[i];
  reverse(s + n / 2 + 1, s + n + 1);
  for (int i = 1; i <= n / 2; i++) ss[i * 2] = s[i + n / 2];
  swap(s, ss);
  int p = 1;
  tot = 1;
  len[1] = -1;
  fail[0] = 1;
  st[0] = 1;
  for (int i = 1; i <= n; i++) {
    p = append(p, i);
    st[i] = p;
  }
  dp[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int p = st[i]; p; p = slink[p]) {
      int cur = dp[i - (len[slink[p]] + dlen[p])];
      gdp[p] = cur;
      if (dlen[p] == dlen[fail[p]]) {
        upd(gdp[p], gdp[fail[p]]);
      }
      if (i % 2 == 0) upd(dp[i], gdp[p]);
    }
  }
  cout << dp[n] << endl;
  return 0;
}