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;
struct state {
  int row, col, val;
  state(int _r, int _c, int _v) : row(_r), col(_c), val(_v) {}
  bool operator<(const state &s) const { return val > s.val; }
};
long long dp[2010];
long long start[2010];
bool palindrome[2010][2010];
string word;
int main() {
  memset(dp, 0, sizeof dp);
  memset(start, 0, sizeof dp);
  cin >> word;
  for (int i = 0; i < word.length() - 1; i++) {
    palindrome[i][i] = 1;
    palindrome[i][i + 1] = (word[i] == word[i + 1]);
  }
  palindrome[word.length() - 1][word.length() - 1] = 1;
  for (int j = 2; j < word.length(); j++) {
    for (int i = 0; i + j < word.length(); i++) {
      palindrome[i][i + j] =
          palindrome[i + 1][j + i - 1] && word[i] == word[i + j];
    }
  }
  for (int i = 0; i < word.length(); i++) {
    for (int j = i; j < word.length(); j++) start[i] += palindrome[i][j];
  }
  dp[word.length() - 1] = 1;
  for (int i = word.length() - 2; i >= 0; i--) {
    dp[i] = dp[i + 1] + start[i];
  }
  long long ans = 0;
  for (int i = 0; i < word.length(); i++) {
    for (int j = i; j < word.length(); j++) {
      if (palindrome[i][j]) ans += dp[j + 1];
    }
  }
  cout << ans << endl;
}