#include <bits/stdc++.h>
char a[2005];
int judge(int l, int r) {
  if (r <= l)
    return 1;
  else {
    if (a[l] == a[r])
      return judge(l + 1, r - 1);
    else
      return 0;
  }
}
int main() {
  long long n, i, j, len, c1[2005], c2[2005];
  scanf("%s", a);
  n = 0;
  len = strlen(a);
  for (i = 1; i <= 2000; i++) {
    c1[i] = 0, c2[i] = 0;
  }
  for (i = 0; i <= len - 1; i++)
    for (j = i; j <= len - 1; j++) {
      if (judge(i, j) == 1) {
        c1[j]++;
        c2[i]++;
      }
    }
  for (i = 0; i <= len - 1; i++)
    for (j = 0; j < i; j++) n = n + c1[j] * c2[i];
  printf("%I64d\n", n);
  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 a[2005];
int judge(int l, int r) {
  if (r <= l)
    return 1;
  else {
    if (a[l] == a[r])
      return judge(l + 1, r - 1);
    else
      return 0;
  }
}
int main() {
  long long n, i, j, len, c1[2005], c2[2005];
  scanf("%s", a);
  n = 0;
  len = strlen(a);
  for (i = 1; i <= 2000; i++) {
    c1[i] = 0, c2[i] = 0;
  }
  for (i = 0; i <= len - 1; i++)
    for (j = i; j <= len - 1; j++) {
      if (judge(i, j) == 1) {
        c1[j]++;
        c2[i]++;
      }
    }
  for (i = 0; i <= len - 1; i++)
    for (j = 0; j < i; j++) n = n + c1[j] * c2[i];
  printf("%I64d\n", n);
  return 0;
}

```