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;
bool vv[2000][2000];
int main() {
  int n;
  string s;
  cin >> s;
  long long left[2000], right[2000];
  for (int i = 0; i < s.length(); i++) {
    vv[i][i] = true;
  }
  int sm = 1, cp = s.length() - 1;
  for (; sm < s.length(); sm++) {
    for (int i = sm, j = 0; i < s.length(); i++, j++) {
      if (s[j] == s[j + sm] && (sm == 1 || vv[j + 1][j + sm - 1])) {
        vv[j][j + sm] = true;
      } else {
        vv[j][j + sm] = false;
      }
    }
  }
  for (int i = 0; i < s.length(); i++) {
    left[i] = 1;
    right[i] = 1;
    for (int j = 0; j < i; j++) {
      if (vv[j][i]) left[i]++;
      if (vv[s.length() - (i + 1)][s.length() - (j + 1)]) right[i]++;
    }
  }
  long long res = 0, right_sum = 0;
  for (int i = 0; i < s.length() - 1; i++) right_sum += right[i];
  for (int i = 0; i < s.length() - 1; i++) {
    res += left[i] * right_sum;
    right_sum -= right[s.length() - (i + 2)];
  }
  cout << res << endl;
  return 0;
}