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;
const int N = 2005;
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int p[N][N];
int le[N], ri[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  string s1;
  string s = " ";
  cin >> s1;
  s += s1;
  int n = s.size();
  n--;
  for (int i = 1; i <= n; i++) p[i][i] = 1;
  for (int l = 2; l <= n; l++) {
    for (int i = 1; i <= n - l + 1; i++) {
      int j = i + l - 1;
      if (l == 2)
        p[i][i + 1] = (s[i] == s[j]);
      else
        p[i][j] = (s[i] == s[j]) && p[i + 1][j - 1];
    }
  }
  for (int j = 1; j <= n; j++) {
    int cnt = 0;
    for (int i = 1; i <= j; i++) {
      if (p[i][j]) cnt++;
    }
    le[j] = cnt;
  }
  for (int j = n; j >= 1; j--) {
    int cnt = 0;
    for (int i = j; i <= n; i++) {
      if (p[j][i]) cnt++;
    }
    ri[j] = ri[j + 1] + cnt;
  }
  long long sum = 0;
  for (int i = 1; i <= n - 1; i++) {
    sum = sum + (le[i] * (1ll) * ri[i + 1]);
  }
  cout << sum << endl;
  return 0;
}