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;
string s;
int n;
bool p[2003][2003], f;
long long hand[2003], tail[2003];
long long ans;
int main() {
  cin >> s;
  n = s.length();
  memset(p, false, sizeof(p));
  for (int i = 0; i <= n - 1; i++) {
    f = true;
    for (int j = 0; j <= i; j++) {
      if (i + j > n) break;
      if (s.substr(i - j, 1) == s.substr(i + j, 1)) {
        p[i - j][i + j] = true;
        hand[i - j]++;
        tail[i + j]++;
      } else {
        f = false;
        break;
      }
    }
  }
  for (int i = 0; i <= n - 1; i++) {
    f = true;
    for (int j = 0; j <= i; j++) {
      if (i + j + 1 > n) break;
      if (s.substr(i - j, 1) == s.substr(i + j + 1, 1)) {
        p[i - j][i + j + 1] = true;
        hand[i - j]++;
        tail[i + j + 1]++;
      } else {
        f = false;
        break;
      }
    }
  }
  for (int i = 1; i <= n - 1; i++) tail[i] += tail[i - 1];
  for (int i = 1; i <= n - 1; i++) ans += hand[i] * tail[i - 1];
  cout << ans << endl;
  return 0;
}