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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s;
  cin >> s;
  long long n = s.length();
  bool palin[n][n];
  memset(palin, false, sizeof palin);
  for (long long i = 0; i < n; i++) palin[i][i] = 1;
  for (long long i = 0; i < n - 1; i++)
    if (s[i] == s[i + 1]) palin[i][i + 1] = 1;
  for (long long len = 2; len < n; len++) {
    for (long long i = 0; i < n - len; i++) {
      long long j = i + len;
      if (s[i] == s[j] && palin[i + 1][j - 1]) palin[i][j] = 1;
    }
  }
  long long sum[n];
  memset(sum, 0, sizeof sum);
  for (long long i = 0; i < n; i++) {
    long long tot = 0;
    for (long long j = 0; j <= i; j++) {
      tot += palin[j][i];
    }
    sum[i] = (i >= 1 ? sum[i - 1] : 0) + tot;
  }
  long long dp[n];
  memset(dp, 0, sizeof dp);
  for (long long i = 0; i < n; i++) {
    long long tot = 0;
    for (long long j = 1; j <= i; j++) {
      tot += sum[j - 1] * palin[j][i];
    }
    dp[i] = (i >= 1 ? dp[i - 1] : 0) + tot;
  }
  cout << dp[n - 1];
  return 0;
}