#include <bits/stdc++.h>
using namespace std;
long long n, lt, rt, pal[2002][2002], st[2002], ed[2002];
string s;
void palindrome(void) {
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      lt = j;
      rt = j + i;
      if (rt >= n) break;
      if (lt == rt)
        pal[lt][rt] = 1;
      else if (lt + 1 == rt && s[lt] == s[rt])
        pal[lt][rt] = 1;
      else if (s[lt] == s[rt] && pal[lt + 1][rt - 1])
        pal[lt][rt] = 1;
    }
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      st[i] += pal[i][j];
      ed[j] += pal[i][j];
    }
  }
  for (long long i = n - 2; i >= 0; i--) st[i] += st[i + 1];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.size();
  palindrome();
  long long ans = 0;
  for (long long i = 0; i < n - 1; i++) ans += ed[i] * st[i + 1];
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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>
using namespace std;
long long n, lt, rt, pal[2002][2002], st[2002], ed[2002];
string s;
void palindrome(void) {
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      lt = j;
      rt = j + i;
      if (rt >= n) break;
      if (lt == rt)
        pal[lt][rt] = 1;
      else if (lt + 1 == rt && s[lt] == s[rt])
        pal[lt][rt] = 1;
      else if (s[lt] == s[rt] && pal[lt + 1][rt - 1])
        pal[lt][rt] = 1;
    }
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      st[i] += pal[i][j];
      ed[j] += pal[i][j];
    }
  }
  for (long long i = n - 2; i >= 0; i--) st[i] += st[i + 1];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.size();
  palindrome();
  long long ans = 0;
  for (long long i = 0; i < n - 1; i++) ans += ed[i] * st[i + 1];
  cout << ans << endl;
  return 0;
}

```