#include <bits/stdc++.h>
char s[10012];
int sum[10012], cnt[10012];
int main() {
  int n, i, j;
  scanf("%s", s);
  n = strlen(s);
  for (i = 0; i < n; ++i)
    for (j = 0; j <= i && s[i - j] == s[i + j]; ++j) {
      cnt[i - j]++;
      sum[i + j]++;
    }
  for (i = 0; i < n - 1; ++i)
    if (s[i] == s[i + 1])
      for (j = 0; j <= i && s[i - j] == s[i + j + 1]; ++j) {
        cnt[i - j]++;
        sum[i + j + 1]++;
      }
  for (i = 1; i < n; ++i) sum[i] += sum[i - 1];
  long long ans = 0;
  for (i = 1; i < n; ++i) ans += sum[i - 1] * (long long)cnt[i];
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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 s[10012];
int sum[10012], cnt[10012];
int main() {
  int n, i, j;
  scanf("%s", s);
  n = strlen(s);
  for (i = 0; i < n; ++i)
    for (j = 0; j <= i && s[i - j] == s[i + j]; ++j) {
      cnt[i - j]++;
      sum[i + j]++;
    }
  for (i = 0; i < n - 1; ++i)
    if (s[i] == s[i + 1])
      for (j = 0; j <= i && s[i - j] == s[i + j + 1]; ++j) {
        cnt[i - j]++;
        sum[i + j + 1]++;
      }
  for (i = 1; i < n; ++i) sum[i] += sum[i - 1];
  long long ans = 0;
  for (i = 1; i < n; ++i) ans += sum[i - 1] * (long long)cnt[i];
  printf("%I64d\n", ans);
  return 0;
}

```