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;
int n;
char s[2003];
int cnte[2003];
int cntb[2003];
int main() {
  while (cin >> s) {
    n = strlen(s);
    memset(cnte, 0, sizeof(int) * n);
    memset(cntb, 0, sizeof(int) * n);
    for (int i = 0; i < n; i++) {
      cnte[i]++;
      cntb[i]++;
      for (int k = 1; i + k < n && i - k >= 0 && s[i + k] == s[i - k]; k++) {
        cnte[i + k]++;
        cntb[i - k]++;
      }
    }
    for (int i = 0; i < n; i++) {
      for (int k = 0; i + k < n && i - k - 1 >= 0 && s[i + k] == s[i - k - 1];
           k++) {
        cnte[i + k]++;
        cntb[i - k - 1]++;
      }
    }
    for (int i = 1; i < n; i++) {
      cnte[i] += cnte[i - 1];
    }
    long long ans = 0;
    for (int i = 0; i < n - 1; i++) {
      ans += cnte[i] * (long long)cntb[i + 1];
    }
    cout << ans << "\n";
  }
  return 0;
}