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;
const int INF = 1 << 29;
const double EPS = 1e-9;
const int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
string str;
int flag[2010][2010];
int sumfrom[2010], sumto[2010];
int determine(int l, int r) {
  if (flag[l][r] == -1) {
    if (r - l <= 0) return flag[l][r] = 1;
    return flag[l][r] = (str[l] == str[r] ? 1 : 0) & determine(l + 1, r - 1);
  }
  return flag[l][r];
}
int main() {
  cin >> str;
  int n = str.size();
  for (int i = 0; i <= n; i++) {
    sumfrom[i] = 0;
    sumto[i] = 0;
    for (int j = 0; j <= n; j++) {
      flag[i][j] = -1;
    }
  }
  for (int i = 0; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      if (determine(i, j)) {
        sumfrom[i]++;
        sumto[j]++;
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      ans += sumto[i] * sumfrom[j];
    }
  }
  cout << ans << endl;
  return 0;
}