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 l[2005], r[2005];
bool dp[2005][2005];
string s;
void all_palindroms(int n) {
  for (int i = 0; i < n; i++) {
    dp[i][i] = true;
    l[i]++;
    r[i]++;
  }
  for (int i = 0; i < n - 1; i++) {
    if (s[i] == s[i + 1]) {
      dp[i][i + 1] = true;
      l[i]++;
      r[i + 1]++;
    }
  }
  for (int gap = 2; gap < n; gap++) {
    for (int i = 0; i < n - gap; i++) {
      int j = gap + i;
      if (s[i] == s[j] && dp[i + 1][j - 1]) {
        dp[i][j] = true;
        l[i]++;
        r[j]++;
      }
    }
  }
}
int main() {
  int n, i, x, j;
  long long sol = 0;
  ios_base::sync_with_stdio(false);
  cin >> s;
  x = s.size();
  all_palindroms(x);
  for (i = 0; i < x; i++) {
    for (j = i + 1; j < x; j++) {
      sol += (long long)r[i] * l[j];
    }
  }
  cout << sol;
  return 0;
}