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>
int main() {
  std::string s;
  std::cin >> s;
  int n = s.size();
  std::vector<std::vector<bool> > matrix(n, std::vector<bool>(n));
  for (int i = 0; i < n; i++) {
    matrix[i][i] = true;
  }
  for (int diff = 1; diff < n; diff++) {
    for (int i = 0; i + diff < n; i++) {
      int j = diff + i;
      if (diff == 1) {
        matrix[i][j] = (s[i] == s[j]);
      } else {
        matrix[i][j] = matrix[i + 1][j - 1] && (s[i] == s[j]);
      }
    }
  }
  std::vector<long long> counts(n);
  counts[0] = 1;
  for (int i = 1; i < n; i++) {
    counts[i] = counts[i - 1];
    for (int j = 0; j < n; j++) {
      if (matrix[j][i]) {
        counts[i]++;
      }
    }
  }
  std::vector<long long> answer(n);
  for (int j = 0; j < n; j++) {
    for (int i = 1; i < n; i++) {
      if (matrix[i][j]) {
        answer[j] += counts[i - 1];
      }
    }
    if (j > 0) {
      answer[j] += answer[j - 1];
    }
  }
  std::cout << answer[n - 1];
}