#include <bits/stdc++.h>
using namespace std;
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tot;
  cin >> tot;
  vector<int> ar(tot);
  unordered_map<int, long long> n, r;
  unordered_set<int> s;
  for (auto &i : ar) {
    cin >> i;
    ++n[i];
  }
  sort(ar.begin(), ar.end());
  long long ans = 1;
  for (int i = 0; i < 3; ++i) {
    ++r[ar[i]];
    s.insert(ar[i]);
  }
  for (auto &i : s) {
    if (r[i] == 2) {
      ans *= (n[i] * (n[i] - 1)) / 2;
    } else if (r[i] == 3) {
      ans *= (n[i] * (n[i] - 1) * (n[i] - 2)) / 6;
    } else {
      ans *= n[i];
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

After returning from the army Makes received a gift — an array a consisting of n positive integer numbers. He hadn't been solving problems for a long time, so he became interested to answer a particular question: how many triples of indices (i, j, k) (i < j < k), such that ai·aj·ak is minimum possible, are there in the array? Help him with it!

Input

The first line of input contains a positive integer number n (3 ≤ n ≤ 105) — the number of elements in array a. The second line contains n positive integer numbers ai (1 ≤ ai ≤ 109) — the elements of a given array.

Output

Print one number — the quantity of triples (i, j, k) such that i, j and k are pairwise distinct and ai·aj·ak is minimum possible.

Examples

Input

4
1 1 1 1


Output

4


Input

5
1 3 2 3 4


Output

2


Input

6
1 3 3 1 3 2


Output

1

Note

In the first example Makes always chooses three ones out of four, and the number of ways to choose them is 4.

In the second example a triple of numbers (1, 2, 3) is chosen (numbers, not indices). Since there are two ways to choose an element 3, then the answer is 2.

In the third example a triple of numbers (1, 1, 2) is chosen, and there's only one way to choose indices.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tot;
  cin >> tot;
  vector<int> ar(tot);
  unordered_map<int, long long> n, r;
  unordered_set<int> s;
  for (auto &i : ar) {
    cin >> i;
    ++n[i];
  }
  sort(ar.begin(), ar.end());
  long long ans = 1;
  for (int i = 0; i < 3; ++i) {
    ++r[ar[i]];
    s.insert(ar[i]);
  }
  for (auto &i : s) {
    if (r[i] == 2) {
      ans *= (n[i] * (n[i] - 1)) / 2;
    } else if (r[i] == 3) {
      ans *= (n[i] * (n[i] - 1) * (n[i] - 2)) / 6;
    } else {
      ans *= n[i];
    }
  }
  cout << ans << '\n';
  return 0;
}

```