#include <bits/stdc++.h>
using namespace std;
long long f(int n, int r) {
  if (r > n) return 0;
  if (r == n || !r) return 1;
  if (r == 3)
    return n * 1LL * (n - 1) * (n - 2) / 6;
  else
    return n * 1LL * (n - 1) / 2;
}
int main() {
  int n;
  cin >> n;
  vector<int> A(n);
  map<int, int> mp;
  set<int> S;
  for (int i = 0; i < n; i++) cin >> A[i], mp[A[i]]++;
  sort(A.begin(), A.end());
  A.resize(3);
  for (int i = 0; i < A.size(); i++) S.insert(A[i]);
  A.clear();
  int sum = 0;
  long long res = 1;
  for (auto no : S) {
    sum += mp[no];
    A.push_back(no);
  }
  if (sum == 3) {
    cout << 1 << endl;
    return 0;
  }
  if (S.size() == 1) {
    cout << f(mp[A[0]], 3) << endl;
  } else if (S.size() == 2) {
    long long ret = 0;
    if (mp[A[0]] == 1)
      ret = f(mp[A[1]], 2);
    else
      ret = mp[A[1]];
    cout << ret << endl;
  } else {
    long long ret = mp[A[0]] * 1LL * mp[A[1]] * mp[A[2]];
    cout << ret << endl;
  }
}

### Prompt

Please create a solution in CPP 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;
long long f(int n, int r) {
  if (r > n) return 0;
  if (r == n || !r) return 1;
  if (r == 3)
    return n * 1LL * (n - 1) * (n - 2) / 6;
  else
    return n * 1LL * (n - 1) / 2;
}
int main() {
  int n;
  cin >> n;
  vector<int> A(n);
  map<int, int> mp;
  set<int> S;
  for (int i = 0; i < n; i++) cin >> A[i], mp[A[i]]++;
  sort(A.begin(), A.end());
  A.resize(3);
  for (int i = 0; i < A.size(); i++) S.insert(A[i]);
  A.clear();
  int sum = 0;
  long long res = 1;
  for (auto no : S) {
    sum += mp[no];
    A.push_back(no);
  }
  if (sum == 3) {
    cout << 1 << endl;
    return 0;
  }
  if (S.size() == 1) {
    cout << f(mp[A[0]], 3) << endl;
  } else if (S.size() == 2) {
    long long ret = 0;
    if (mp[A[0]] == 1)
      ret = f(mp[A[1]], 2);
    else
      ret = mp[A[1]];
    cout << ret << endl;
  } else {
    long long ret = mp[A[0]] * 1LL * mp[A[1]] * mp[A[2]];
    cout << ret << endl;
  }
}

```