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 main() {
  string s;
  cin >> s;
  int n = s.size();
  long long ans = 0;
  int psum[n], no_of_p_start_at[n], no_of_p_end_at[n];
  memset(psum, 0, sizeof(psum));
  memset(no_of_p_start_at, 0, sizeof(no_of_p_start_at));
  memset(no_of_p_end_at, 0, sizeof(no_of_p_end_at));
  for (int i = 0; i < n; i++) {
    int j = 0;
    while ((i - j) >= 0 && (i + j) < n) {
      if (s[i - j] == s[i + j]) {
        no_of_p_start_at[i - j]++;
        no_of_p_end_at[i + j]++;
        j++;
        continue;
      } else
        break;
    }
    int l = i - 1;
    int r = i;
    while ((l) >= 0 && (r) < n) {
      if (s[l] == s[r]) {
        no_of_p_start_at[l]++;
        no_of_p_end_at[r]++;
        l--;
        r++;
        continue;
      } else
        break;
    }
  }
  psum[0] = 1;
  for (int i = 1; i < n; i++) {
    psum[i] = psum[i - 1] + no_of_p_end_at[i];
  }
  for (int i = 1; i < n; i++) {
    ans += psum[i - 1] * no_of_p_start_at[i];
  }
  cout << ans << endl;
  cin >> ans;
  return 0;
}