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;
const long long mod = 1e9 + 7;
using namespace std;
string s;
long long ispalindroma[2005][2005];
long long ispalindrome(long long i, long long j) {
  if (i > j) return (1);
  if (i == j) return (1);
  if (ispalindroma[i][j] != -1) return (ispalindroma[i][j]);
  long long ans = 0;
  if (s[i] == s[j]) {
    ans = ispalindrome(i + 1, j - 1);
  }
  return (ispalindroma[i][j] = ans);
}
int solve() {
  cin >> s;
  long long sum[2005], dp[2005];
  dp[0] = sum[0] = 0;
  for (long long i = 0; i < 2005; i++) {
    for (long long j = 0; j < 2005; j++) ispalindroma[i][j] = -1;
  }
  long long a1 = ispalindrome(0, s.size() - 1);
  for (long long i = 0; i < s.size(); i++) {
    long long c = 0;
    for (long long j = 0; j <= i; j++) {
      if (ispalindrome(j, i)) c++;
    }
    sum[i + 1] = sum[i] + c;
  }
  for (long long i = 0; i < s.size(); i++) {
    long long ans = 0;
    for (long long j = 0; j <= i; j++) {
      if (ispalindrome(j, i)) ans = ans + sum[j];
    }
    dp[i + 1] = dp[i] + ans;
  }
  cout << dp[s.size()] << "\n";
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
}