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 pct(int x) { return __builtin_popcount(x); }
int pct(long long x) { return __builtin_popcountll(x); }
int bt(int x) { return 31 - __builtin_clz(x); }
int bt(long long x) { return 63 - __builtin_clzll(x); }
int cdiv(int a, int b) { return a / b + !(a < 0 || a % b == 0); }
long long cdiv(long long a, long long b) {
  return a / b + !(a < 0 || a % b == 0);
}
int nxt_C(int x) {
  int c = x & -x, r = x + c;
  return (((r ^ x) >> 2) / c) | r;
}
long long nxt_C(long long x) {
  long long c = x & -x, r = x + c;
  return (((r ^ x) >> 2) / c) | r;
}
vector<int> get_bits(int mask) {
  vector<int> bb;
  while (mask) {
    int b = bt(mask);
    bb.push_back(b);
    mask ^= (1 << b);
  }
  reverse(bb.begin(), bb.end());
  return bb;
}
int get_mask(vector<int> v) {
  int mask = 0;
  for (int x : v) {
    mask ^= (1 << x);
  }
  return mask;
}
template <typename T>
void uniq(vector<T> &v) {
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
}
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand(long long l, long long r) {
  uniform_int_distribution<long long> uid(l, r);
  return uid(rng);
}
void sc() {}
template <typename Head, typename... Tail>
void sc(Head &H, Tail &...T) {
  cin >> H;
  sc(T...);
}
const int mod = 1e9 + 7;
int pwr(int a, long long b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * 1LL * a) % mod;
    a = (a * 1LL * a) % mod;
    b >>= 1;
  }
  return ans;
}
const int maxn = 1e6 + 5, sigma = 26;
int len[maxn], link[maxn], to[maxn][sigma];
int slink[maxn], diff[maxn], series_ans[maxn];
int sz, last, n;
char s[maxn];
void init() {
  s[n++] = -1;
  link[0] = 1;
  len[1] = -1;
  sz = 2;
}
int get_link(int v) {
  while (s[n - len[v] - 2] != s[n - 1]) v = link[v];
  return v;
}
void add_letter(char c) {
  s[n++] = c -= 'a';
  last = get_link(last);
  if (!to[last][c]) {
    len[sz] = len[last] + 2;
    link[sz] = to[get_link(link[last])][c];
    diff[sz] = len[sz] - len[link[sz]];
    if (diff[sz] == diff[link[sz]])
      slink[sz] = slink[link[sz]];
    else
      slink[sz] = link[sz];
    to[last][c] = sz++;
  }
  last = to[last][c];
}
const int N = 1e6 + 5;
int ans[N];
void solve() {
  init();
  string t;
  cin >> t;
  string s = "";
  for (int i = 0; i <= (int)t.size() / 2 - 1; i++) {
    s += t[i];
    s += t[(int)t.size() - 1 - i];
  }
  int n = s.size();
  ans[0] = 1;
  for (int i = 1; i <= n; i++) {
    add_letter(s[i - 1]);
    for (int v = last; len[v] > 0; v = slink[v]) {
      series_ans[v] = ans[i - (len[slink[v]] + diff[v])];
      if (diff[v] == diff[link[v]]) {
        series_ans[v] += series_ans[link[v]];
        series_ans[v] %= mod;
      }
      ans[i] += series_ans[v];
      ans[i] %= mod;
    }
    if (i & 1) ans[i] = 0;
  }
  cout << ans[n];
}
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int tt = 1; tt <= t; tt++) {
    solve();
  }
  return 0;
}