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>
const double pi = 4.0 * atan(1.0);
const double e = exp(1.0);
const int maxn = 1e6 + 8;
const long long mod = 1e9 + 7;
using namespace std;
long long s[2008][2008];
char a[2008];
int main() {
  scanf("%s", a + 1);
  int l = strlen(a + 1);
  for (int i = 1; i <= l; i++) {
    s[i][i] = 1;
  }
  for (int i = l; i >= 1; i--) {
    for (int j = i + 1; j <= l; j++) {
      if (a[i] == a[j]) {
        if (j == i + 1 || s[i + 1][j - 1] == 1) {
          s[i][j] = 1;
        }
      }
    }
  }
  for (int i = 1; i <= l; i++) {
    for (int k = i + 1; k <= l; k++) {
      s[i][k] += s[i][k - 1];
    }
  }
  long long ans = 0;
  for (int i = 1; i <= l; i++) {
    for (int k = i; k <= l; k++) {
      ans += s[i][k] * s[k + 1][l];
    }
  }
  cout << ans << endl;
}