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;
void disp(long long int i, long long int j, string s);
int main() {
  string s;
  cin >> s;
  bool dp[s.size()][s.size()];
  long long int f = 0;
  long long int counter1[s.size()];
  fill(counter1, counter1 + s.size(), 1);
  for (long long int i = 0; i < s.size(); i++) {
    dp[i][i] = true;
  }
  for (long long int l = 2; l <= s.size(); l++) {
    for (long long int i = 0; i <= s.size() - l; i++) {
      f = i + l - 1;
      if (f == i + 1) {
        dp[i][f] = (s[i] == s[f]);
      } else {
        dp[i][f] = dp[i + 1][f - 1] && (s[i] == s[f]);
      }
      if (dp[i][f]) {
        counter1[i]++;
      }
    }
  }
  for (long long int i = s.size() - 2; i >= 0; i--) {
    counter1[i] += counter1[i + 1];
  }
  long long int c = 0;
  for (long long int i = 0; i < s.size(); i++) {
    for (long long int j = i; j < s.size(); j++) {
      if (dp[i][j]) {
        long long int c1 = 0;
        if (j + 1 < s.size()) {
          c1 = counter1[j + 1];
        }
        c += c1;
      }
    }
  }
  cout << c;
}
void disp(long long int i, long long int j, string s) {
  for (long long int k = i; k <= j; k++) {
    cout << s[k];
  }
}