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 fmo(int x) { return x + ((x >> 31) & 1000000007); }
int n;
char str[1000005], s[1000005];
struct node {
  int len, dif, f;
  node *fa, *top, *ch[26];
} t[1000005], *rt = t, *lst = t;
int tot;
inline node *nw(int len) {
  node *p = t + (++tot);
  p->len = len;
  p->fa = p->top = rt;
  for (int i = 0; i < 26; i++) p->ch[i] = rt;
  return p;
}
inline node *gfa(node *p, int pos) {
  while (s[pos] != s[pos - p->len - 1]) p = p->fa;
  return p;
}
inline void init() {
  rt->top = rt;
  rt->fa = t + 1;
  for (int i = 0; i < 26; i++) rt->ch[i] = rt;
  nw(-1);
}
inline void ins(int c, int pos) {
  node *p = gfa(lst, pos);
  if (p->ch[c] == rt) {
    node *q = nw(p->len + 2);
    q->fa = gfa(p->fa, pos)->ch[c];
    p->ch[c] = q;
    q->dif = q->len - q->fa->len;
    q->top = q->dif == q->fa->dif ? q->fa->top : q->fa;
  }
  lst = p->ch[c];
}
int f[1000005];
int main() {
  scanf("%s", str + 1);
  n = strlen(str + 1);
  for (int i = 1, l = 1, r = n; i <= n; i++) s[i] = str[(i & 1) ? l++ : r--];
  init();
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    ins(s[i] - 'a', i);
    for (node *p = lst; p != rt; p = p->top) {
      p->f = f[i - p->top->len - p->dif];
      if (p->top != p->fa) p->f = fmo(p->f + p->fa->f - 1000000007);
      if (i % 2 == 0) f[i] = fmo(f[i] + p->f - 1000000007);
    }
  }
  printf("%d\n", f[n]);
}