#include <bits/stdc++.h>
using namespace std;
const long double pi = 3.1415926535897932384626433;
const int mod = 1000000007;
const long long inf = 3e18;
long long ncr(long long n, long long r) {
  if (r == 0) {
    return 1;
  }
  if (r == 1) {
    return n;
  }
  if (r == 2) {
    return n * (n - 1) / 2;
  }
  if (r == 3) return n * (n - 1) * (n - 2) / 6;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  vector<long long> a(n);
  map<long long, long long> mp;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]]++;
  }
  sort(a.begin(), a.end());
  long long ans;
  if (mp[a[0]] >= 3) {
    ans = ncr(mp[a[0]], 3);
  } else if (mp[a[0]] + mp[a[1]] >= 3) {
    ans = ncr(mp[a[0 + mp[a[0]]]], 3 - mp[a[0]]);
  } else {
    ans = mp[a[2]];
  }
  cout << ans;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const long double pi = 3.1415926535897932384626433;
const int mod = 1000000007;
const long long inf = 3e18;
long long ncr(long long n, long long r) {
  if (r == 0) {
    return 1;
  }
  if (r == 1) {
    return n;
  }
  if (r == 2) {
    return n * (n - 1) / 2;
  }
  if (r == 3) return n * (n - 1) * (n - 2) / 6;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  vector<long long> a(n);
  map<long long, long long> mp;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]]++;
  }
  sort(a.begin(), a.end());
  long long ans;
  if (mp[a[0]] >= 3) {
    ans = ncr(mp[a[0]], 3);
  } else if (mp[a[0]] + mp[a[1]] >= 3) {
    ans = ncr(mp[a[0 + mp[a[0]]]], 3 - mp[a[0]]);
  } else {
    ans = mp[a[2]];
  }
  cout << ans;
}

```