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 int B[101010], C[101010];
int main() {
  long long int i, j, k, count = 0;
  string s;
  cin >> s;
  for (i = 0; i < s.length(); i++) {
    B[i] += 1;
    C[s.length()] += 1;
    C[i] -= 1;
    j = i + 1;
    k = i - 1;
    while (j < s.length() && k >= 0) {
      if (s[j] != s[k]) break;
      B[k] += 1;
      C[s.length()] += 1;
      C[j] -= 1;
      j++;
      k--;
    }
    k = i;
    j = i + 1;
    while (j < s.length() && k >= 0) {
      if (s[j] != s[k]) break;
      B[k] += 1;
      C[s.length()] += 1;
      C[j] -= 1;
      j++;
      k--;
    }
  }
  for (i = s.length() - 1; i >= 0; i--) C[i] = C[i + 1] + C[i];
  for (i = 0; i < s.length(); i++) count += (C[i]) * (B[i]);
  cout << count << endl;
  return (0);
}