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;
char s[2002], sum[2002];
long long pal1[2002], pal2[2002];
int Min(int a, int b) {
  if (a < b) return a;
  return b;
}
int Max(int a, int b) {
  if (a > b) return a;
  return b;
}
int main() {
  int i, j, len;
  long long res;
  scanf("%s", s);
  len = strlen(s);
  memset(pal1, 0, sizeof(pal1));
  memset(pal2, 0, sizeof(pal2));
  sum[0] = 0;
  for (i = 1; i < 2002; i++) {
    sum[i] = sum[i - 1] + i;
  }
  for (i = 0; i < len; i++) {
    for (j = 0; j < len; j++) {
      if (i - j < 0 || i + j >= len)
        break;
      else {
        if (s[i + j] == s[i - j]) {
          pal1[i - j]++;
          pal2[i + j]++;
        } else
          break;
      }
    }
    for (j = 0; j < len; j++) {
      if (i - j < 0 || i + 1 + j >= len)
        break;
      else {
        if (s[i - j] == s[i + 1 + j]) {
          pal1[i - j]++;
          pal2[i + 1 + j]++;
        } else
          break;
      }
    }
  }
  for (i = len - 2; i >= 0; i--) pal1[i] += pal1[i + 1];
  res = 0;
  for (i = 0; i < len - 1; i++) {
    res += (pal2[i] * pal1[i + 1]);
  }
  cout << res << endl;
  return 0;
}