#include <bits/stdc++.h>
using namespace std;
map<int, int> hh;
vector<pair<long long, long long> > v;
int main() {
  int n;
  cin >> n;
  int i, j, z;
  for (i = 0; i < n; i++) cin >> z, hh[z]++;
  for (auto itr = hh.begin(); itr != hh.end() && (int)(v).size() < 3; itr++)
    v.push_back({1LL * (*itr).first, 1LL * (*itr).second});
  long long ans = 0;
  if (v[0].second >= 3)
    ans = v[0].second * (v[0].second - 1) * (v[0].second - 2) / 6;
  else {
    if (v[0].second == 2)
      ans = v[0].second * (v[0].second - 1) / 2 * v[1].second;
    else if (v[0].second == 1 && v[1].second >= 2)
      ans = v[1].second * (v[1].second - 1) / 2;
    else
      ans = v[2].second;
  }
  cout << ans;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
map<int, int> hh;
vector<pair<long long, long long> > v;
int main() {
  int n;
  cin >> n;
  int i, j, z;
  for (i = 0; i < n; i++) cin >> z, hh[z]++;
  for (auto itr = hh.begin(); itr != hh.end() && (int)(v).size() < 3; itr++)
    v.push_back({1LL * (*itr).first, 1LL * (*itr).second});
  long long ans = 0;
  if (v[0].second >= 3)
    ans = v[0].second * (v[0].second - 1) * (v[0].second - 2) / 6;
  else {
    if (v[0].second == 2)
      ans = v[0].second * (v[0].second - 1) / 2 * v[1].second;
    else if (v[0].second == 1 && v[1].second >= 2)
      ans = v[1].second * (v[1].second - 1) / 2;
    else
      ans = v[2].second;
  }
  cout << ans;
  return 0;
}

```