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.length();
  int palin[n][n];
  memset(palin, 0, sizeof(palin));
  for (int i = 0; i < n; i++) {
    palin[i][i] = 1;
    if (s[i] == s[i + 1]) palin[i][i + 1] = 1;
  }
  for (int i = 2; i < n; i++)
    for (int j = 0; j < n - i; j++)
      if (s[j] == s[j + i]) palin[j][j + i] = palin[j + 1][j + i - 1];
  long long int sum[n];
  memset(sum, 0, sizeof(sum));
  sum[0] = 1;
  for (int i = 1; i < n; i++) {
    sum[i] = sum[i - 1];
    for (int j = 0; j <= i; j++) sum[i] += palin[j][i];
  }
  long long int dp[n];
  memset(dp, 0, sizeof(dp));
  dp[1] = 1;
  for (int i = 2; i < n; i++) {
    dp[i] += dp[i - 1];
    for (int j = 0; j <= i - 1; j++)
      if (palin[j + 1][i] == 1) dp[i] += sum[j];
  }
  cout << dp[n - 1] << endl;
  return 0;
}