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 is_palin[2010][2010];
string s;
int dp1[2010], dp2[2010];
long long int solve(int start, int end) {
  if (is_palin[start][end] != -1) return is_palin[start][end];
  if (start > end)
    return -1;
  else {
    if (s[start] == s[end]) {
      is_palin[start][end] = solve(start + 1, end - 1);
    } else
      is_palin[start][end] = -1;
    return is_palin[start][end];
  }
}
int main() {
  cin >> s;
  memset(is_palin, -1, sizeof is_palin);
  int len = s.length();
  for (int i = 0; i < len; i++) is_palin[i][i] = 1;
  for (int i = 0; i < len - 1; i++)
    if (s[i] == s[i + 1]) is_palin[i][i + 1] = 1;
  solve(0, len - 1);
  for (int i = 0; i < len; i++) {
    for (int j = i; j < len; j++) {
      is_palin[i][j] = solve(i, j);
    }
  }
  long long int ans = 0;
  dp1[0] = 1;
  for (int i = 1; i < len; i++) {
    for (int j = 0; j <= i; j++) {
      if (is_palin[j][i] == 1) dp1[i]++;
    }
  }
  dp2[len - 1] = 1;
  for (int i = len - 2; i >= 0; i--) {
    dp2[i] = dp2[i + 1];
    for (int j = len - 1; j >= i; j--) {
      if (is_palin[i][j] == 1) dp2[i]++;
    }
  }
  for (int i = 0; i < len - 1; i++) {
    ans += dp1[i] * dp2[i + 1];
  }
  cout << ans << endl;
}