#include <bits/stdc++.h>
using namespace std;
long long memo[2005];
vector<long long> palind[2005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  long long n, i, j, ans = 0;
  string s;
  cin >> s;
  s = '$' + s;
  n = s.size();
  for (i = 2; i < n; i++) {
    memo[i] = memo[i - 1];
    if (s[i] == s[i - 1]) {
      palind[i].push_back(i - 1);
      memo[i]++;
      ans += i - 2 + n - i - 1;
      ans += memo[i - 2];
    }
    if (s[i - 2] == s[i]) {
      palind[i].push_back(i - 2);
      memo[i]++;
      ans += i - 3 + n - i - 1;
      ans += memo[i - 3];
    }
    for (auto p : palind[i - 1]) {
      if (s[i] == s[p - 1]) {
        palind[i].push_back(p - 1);
        memo[i]++;
        ans += p - 2 + n - i - 1;
        ans += memo[p - 2];
      }
    }
  }
  cout << ans + (n - 1) * (n - 2) / 2 << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 memo[2005];
vector<long long> palind[2005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  long long n, i, j, ans = 0;
  string s;
  cin >> s;
  s = '$' + s;
  n = s.size();
  for (i = 2; i < n; i++) {
    memo[i] = memo[i - 1];
    if (s[i] == s[i - 1]) {
      palind[i].push_back(i - 1);
      memo[i]++;
      ans += i - 2 + n - i - 1;
      ans += memo[i - 2];
    }
    if (s[i - 2] == s[i]) {
      palind[i].push_back(i - 2);
      memo[i]++;
      ans += i - 3 + n - i - 1;
      ans += memo[i - 3];
    }
    for (auto p : palind[i - 1]) {
      if (s[i] == s[p - 1]) {
        palind[i].push_back(p - 1);
        memo[i]++;
        ans += p - 2 + n - i - 1;
        ans += memo[p - 2];
      }
    }
  }
  cout << ans + (n - 1) * (n - 2) / 2 << endl;
  return 0;
}

```