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 a[2005][2005];
int64_t dpt[2005];
void dp(int n, string s) {
  for (int i = 0; i < n; i++) {
    a[i][i] = 1;
  }
  for (int l = 2; l <= n; l++) {
    for (int i = 0; i < n - l + 1; i++) {
      if (s[i] == s[i + l - 1]) {
        if (l == 2 || a[i + 1][i + l - 2]) {
          a[i][i + l - 1] = 1;
        }
      }
    }
  }
}
int main() {
  string s;
  cin >> s;
  int n = s.length();
  dp(n, s);
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      if (a[i - 1][j - 1]) {
        dpt[j]++;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    dpt[i] += dpt[i - 1];
  }
  int64_t ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      if (a[i - 1][j - 1]) {
        ans += dpt[i - 1];
      }
    }
  }
  cout << ans;
}