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;
long long isPal[2005][2005];
long long ans, N;
long long dp[2005][2005];
string a;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> a;
  N = a.size() - 1;
  for (long long i = 0; i <= N; i++) {
    dp[i][i] = 1;
    isPal[i][i] = 1;
  }
  for (long long t = 2; t <= N; t++)
    for (long long i = 0; i + t - 1 <= N; i++) {
      long long l = i, r = l + t - 1;
      if (a[l] == a[r] && (t == 2 or isPal[l + 1][r - 1])) isPal[l][r] = 1;
      dp[l][r] = dp[l][r - 1] + dp[l + 1][r] - dp[l + 1][r - 1] + isPal[l][r];
    }
  for (long long i = 0; i <= N; i++)
    for (long long j = i; j <= N; j++)
      if (isPal[i][j]) {
        if (i > 0) ans += dp[0][i - 1];
        if (j < N) ans += dp[j + 1][N];
      }
  cout << ans / 2 << "\n";
  return 0;
}