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 can[2100][2100];
long long best[2100][2100];
char s[2100];
int main() {
  int n, i, j;
  scanf("%s", s);
  n = strlen(s);
  memset(can, 0, sizeof(can));
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      if (i - j < 0 || i + j >= n) break;
      if (s[i - j] != s[i + j]) break;
      can[i - j][i + j] = 1;
    }
    for (j = 0; j < n; j++) {
      if (i - j < 0 || i + j + 1 >= n) break;
      if (s[i - j] != s[i + j + 1]) break;
      can[i - j][i + j + 1] = 1;
    }
  }
  memset(best, 0, sizeof(best));
  for (best[0][0] = i = 1; i < n; i++) {
    best[i][0] = best[i - 1][0];
    for (j = 0; j <= i; j++)
      if (can[j][i]) best[i][0]++;
  }
  for (i = 1; i < n; i++) {
    best[i][1] = best[i - 1][1];
    for (j = 1; j <= i; j++)
      if (can[j][i]) best[i][1] += best[j - 1][0];
  }
  printf("%I64d\n", best[n - 1][1]);
  return 0;
}