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;
string s;
int n;
long long f[2002][2002], res;
int pl[2002][2002];
int main() {
  cin >> s;
  n = s.length();
  res = 0;
  int j;
  memset(pl, 0, sizeof(pl));
  for (int i = 0; i <= n - 1; i++) pl[i][i] = 1;
  for (int i = 0; i <= n - 2; i++)
    if (s[i] == s[i + 1]) pl[i][i + 1] = 1;
  for (int k = 2; k <= n - 1; k++)
    for (int i = 0; i <= n - 1 - k; i++) {
      j = i + k;
      if (s[i] == s[j])
        pl[i][j] = pl[i + 1][j - 1];
      else
        pl[i][j] = 0;
    }
  memset(f, 0, sizeof(f));
  for (int i = 0; i <= n - 1; i++) f[i][i] = 1;
  for (int i = 0; i <= n - 2; i++)
    if (s[i] == s[i + 1])
      f[i][i + 1] = 3;
    else
      f[i][i + 1] = 2;
  for (int k = 2; k <= n - 1; k++)
    for (int i = 0; i <= n - 1 - k; i++) {
      j = i + k;
      f[i][j] = f[i + 1][j] + f[i][j - 1] - f[i + 1][j - 1];
      if (pl[i][j]) f[i][j]++;
    }
  for (int i = 0; i <= n - 2; i++)
    for (int j = i; j <= n - 2; j++)
      if (pl[i][j]) res += f[j + 1][n - 1];
  cout << res;
  return 0;
}