#include <bits/stdc++.h>
char str[2009];
long long int startp[2009];
long long int endp[2009];
int isPal(int a, int b) {
  while (a <= b) {
    if (str[a] != str[b]) return 0;
    a++;
    b--;
  }
  return 1;
}
int main() {
  int n, i, j;
  scanf("%s", str);
  n = strlen(str);
  for (i = 0; i < n; i++) {
    startp[i] = 1;
    endp[i] = 1;
  }
  long long int ans = 0;
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      if (isPal(i, j)) {
        startp[i]++;
        endp[j]++;
      }
    }
    if (i > 0) {
      ans += (endp[i - 1] * startp[i]);
      endp[i] += endp[i - 1];
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
char str[2009];
long long int startp[2009];
long long int endp[2009];
int isPal(int a, int b) {
  while (a <= b) {
    if (str[a] != str[b]) return 0;
    a++;
    b--;
  }
  return 1;
}
int main() {
  int n, i, j;
  scanf("%s", str);
  n = strlen(str);
  for (i = 0; i < n; i++) {
    startp[i] = 1;
    endp[i] = 1;
  }
  long long int ans = 0;
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      if (isPal(i, j)) {
        startp[i]++;
        endp[j]++;
      }
    }
    if (i > 0) {
      ans += (endp[i - 1] * startp[i]);
      endp[i] += endp[i - 1];
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```