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;
long long dp[2003][2003];
long long dp2[2003];
long long dp3[2003];
char a[2003];
int main() {
  cin >> a;
  long long ans = 0;
  int len;
  int l, r;
  int i, j;
  for (len = 0; a[len]; len++)
    ;
  for (i = 0; i < len; i++) {
    l = i - 1;
    r = i + 1;
    while (l > -1 && r < len && a[l] == a[r]) {
      dp[l][r] = 1;
      l--;
      r++;
    }
    if (a[i - 1] == a[i] && i - 1 > -1) {
      dp[i - 1][i] = 1;
      l = i - 2;
      r = i + 1;
      while (l > -1 && r < len && a[r] == a[l]) {
        dp[l][r] = 1;
        l--;
        r++;
      }
    }
  }
  dp2[0] = 1;
  int res;
  for (i = 1; i < len; i++) {
    res = 0;
    for (j = i - 1; j > -1; j--) res += dp[j][i];
    dp2[i] = dp2[i - 1] + res + 1;
  }
  dp3[len - 1] = 1;
  for (i = len - 2; i > -1; i--) {
    res = 0;
    for (j = i + 1; j < len; j++) res += dp[i][j];
    dp3[i] = dp3[i + 1] + res + 1;
  }
  for (i = 1; i + 1 < len; i++) ans += (dp2[i] - dp2[i - 1]) * dp3[i + 1];
  cout << ans + dp3[1] << endl;
}