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;
bool dp[2001][2001];
int main() {
  string s;
  cin >> s;
  memset(dp, false, sizeof(dp));
  long long l[s.size() + 1];
  long long r[s.size() + 1];
  for (int i = 0; i < s.size(); i++) {
    dp[i][i] = true;
    l[i] = 1;
    r[i] = 1;
  }
  for (int l1 = 2; l1 <= s.size(); l1++) {
    for (int i = 0; i <= s.size() - l1; i++) {
      int j = i + l1 - 1;
      if (s[i] == s[j]) {
        if (i + 1 != j)
          dp[i][j] = dp[i + 1][j - 1];
        else
          dp[i][j] = true;
        if (dp[i][j]) {
          l[j]++;
          r[i]++;
        }
      }
    }
  }
  for (int j = 1; j < s.size(); j++) {
    l[j] = l[j - 1] + l[j];
  }
  long long answ = 0;
  for (int k = 1; k <= s.size() - 1; k++) {
    answ = answ + (l[k - 1] * r[k]);
  }
  cout << answ << endl;
}