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 N = 1000010;
const int mod = 1e9 + 7;
template <typename Tp>
void Add(Tp &x, Tp y) {
  x = (x + y) % mod;
}
namespace PAM {
const int MAX = 1000010;
int tr[MAX][26], fail[MAX], len[MAX], diff[MAX], slink[MAX];
int sz, tot, lst;
char s[MAX];
int newnode(int l) {
  ++sz;
  len[sz] = l;
  fail[sz] = diff[sz] = slink[sz] = 0;
  memset(tr[sz], 0, sizeof(tr[sz]));
  return sz;
}
void init() {
  sz = -1;
  lst = 0;
  s[tot = 0] = '$';
  newnode(0);
  newnode(-1);
  fail[0] = 1;
}
int getfail(int x) {
  while (s[tot] != s[tot - len[x] - 1]) x = fail[x];
  return x;
}
void insert(char c) {
  s[++tot] = c;
  int cur = getfail(lst);
  if (!tr[cur][c - 'a']) {
    int now = newnode(len[cur] + 2);
    fail[now] = tr[getfail(fail[cur])][c - 'a'];
    tr[cur][c - 'a'] = now;
    diff[now] = len[now] - len[fail[now]];
    if (diff[now] == diff[fail[now]])
      slink[now] = slink[fail[now]];
    else
      slink[now] = fail[now];
  }
  lst = tr[cur][c - 'a'];
}
long long solve(char *s, int n) {
  static long long g[N], dp[N];
  init();
  memset(g, 0, sizeof(g));
  memset(dp, 0, sizeof(dp));
  dp[0] = 1;
  for (int i = 1; i <= n; ++i) {
    insert(s[i]);
    for (int x = lst; x > 1; x = slink[x]) {
      g[x] = dp[i - len[slink[x]] - diff[x]];
      if (diff[x] == diff[fail[x]]) Add(g[x], g[fail[x]]);
      if (i % 2 == 0) Add(dp[i], g[x]);
    }
  }
  return dp[n];
}
}  // namespace PAM
char s[N], t[N];
int n;
int main() {
  cin >> (s + 1);
  n = strlen(s + 1);
  if (n & 1) {
    puts("0");
    return 0;
  }
  for (int i = 1; i <= n / 2; ++i) {
    t[i * 2 - 1] = s[i];
    t[i * 2] = s[n - i + 1];
  }
  cout << PAM::solve(t, n) << endl;
  return 0;
}