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;
char pal[2005][2005];
int shere[2005];
int here_on[2005];
void genPal() {
  int n = s.size();
  for (int i = 0; i < (n); i++)
    for (int j = (0); j <= (i); j++) pal[i][j] = 1;
  for (int tam = (1); tam <= (n - 1); tam++)
    for (int i = 0; i < (n); i++) {
      int j = i + tam;
      if (j >= n) break;
      if (s[i] != s[j])
        pal[i][j] = 0;
      else
        pal[i][j] = pal[i + 1][j - 1];
    }
  if (false) {
    for (int i = 0; i < (n); i++) {
      for (int j = 0; j < (n); j++) printf("%d,", pal[i][j]);
      printf("\n");
    }
  }
}
int main() {
  cin >> s;
  int n = s.size();
  genPal();
  long long ans = 0;
  for (int i = 0; i < (n); i++) {
    shere[i] = 0;
    for (int j = (i); j <= (n); j++) shere[i] += pal[i][j];
  }
  here_on[n] = 0;
  for (int i = (n - 1); i >= (0); i--) {
    here_on[i] = here_on[i + 1] + shere[i];
  }
  for (int i = 0; i < (n); i++) {
    for (int j = (i); j <= (n - 1); j++) {
      if (pal[i][j]) ans += here_on[j + 1];
    }
  }
  cout << ans << "\n";
}