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;
string arr;
int dp[2005][2005];
int solv(int i, int j) {
  if (i >= j) return dp[i][j] = 1;
  if (dp[i][j] != -1) return dp[i][j];
  if (arr[i] == arr[j]) return dp[i][j] = solv(i + 1, j - 1);
  return dp[i][j] = 0;
}
int main() {
  cin >> arr;
  memset(dp, -1, sizeof dp);
  for (int i = 0; i < arr.size(); i++) {
    for (int j = i; j < arr.size(); j++) solv(i, j);
  }
  long long acc[arr.size() + 1];
  acc[arr.size()] = 0;
  for (int i = arr.size() - 1; i >= 0; i--) {
    acc[i] = acc[i + 1];
    for (int j = i; j < arr.size(); j++) acc[i] += dp[i][j];
  }
  long long ans = 0;
  for (int i = 0; i < arr.size(); i++) {
    for (int j = i; j < arr.size(); j++) {
      if (dp[i][j] == 1) ans += acc[j + 1];
    }
  }
  cout << ans << endl;
  return 0;
}