#include <bits/stdc++.h>
using namespace std;
long long n, i, j, d1[2000], d2[2000], sum[5000], ans;
string s;
int main() {
  cin >> s;
  n = s.length();
  for (i = 0; i < n; i++) {
    j = 0;
    while (i - j >= 0 && i + j < n && s[i - j] == s[i + j]) {
      d1[i - j]++;
      d2[i + j]++;
      j++;
    }
    j = 0;
    while (i - j - 1 >= 0 && i + j < n && s[i - j - 1] == s[i + j]) {
      d1[i - j - 1]++;
      d2[i + j]++;
      j++;
    }
  }
  for (i = n - 1; i >= 0; i--) sum[i] = sum[i + 1] + d1[i];
  for (i = 0; i < n; i++) ans += d2[i] * sum[i + 1];
  cout << ans;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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>
using namespace std;
long long n, i, j, d1[2000], d2[2000], sum[5000], ans;
string s;
int main() {
  cin >> s;
  n = s.length();
  for (i = 0; i < n; i++) {
    j = 0;
    while (i - j >= 0 && i + j < n && s[i - j] == s[i + j]) {
      d1[i - j]++;
      d2[i + j]++;
      j++;
    }
    j = 0;
    while (i - j - 1 >= 0 && i + j < n && s[i - j - 1] == s[i + j]) {
      d1[i - j - 1]++;
      d2[i + j]++;
      j++;
    }
  }
  for (i = n - 1; i >= 0; i--) sum[i] = sum[i + 1] + d1[i];
  for (i = 0; i < n; i++) ans += d2[i] * sum[i + 1];
  cout << ans;
  return 0;
}

```