Problem: You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.

In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.

A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.

A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".

Input

The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.

Output

Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.

Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.

Examples

Input

aa


Output

1


Input

aaa


Output

5


Input

abacaba


Output

36

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int power(long long int x, unsigned long long y, int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    string s;
    cin >> s;
    int n = s.length();
    vector<int> sufs(n, 0), sufe(n, 0);
    for (int i = 0; i < n; i++) {
      for (int j = i, k = i; (k > -1) && (j < n); j++, k--) {
        if (s[j] == s[k]) {
          sufe[k]++;
          sufs[j]++;
        } else
          break;
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = i, k = i + 1; (j > -1) && (k < n); j--, k++) {
        if (s[j] == s[k]) {
          sufs[k]++;
          sufe[j]++;
        } else
          break;
      }
    }
    for (int i = n - 2; i > -1; i--) {
      sufe[i] += sufe[i + 1];
    }
    long long int ans = 0;
    for (int i = 0; i < n - 1; i++) {
      ans += ((long long int)sufs[i]) * ((long long int)sufe[i + 1]);
    }
    cout << ans << endl;
  }
  return 0;
}