#include <bits/stdc++.h>
using namespace std;
long long mins[3];
long long freqs[3];
void move(int i) {
  for (int j = 2; j > i; j--) {
    mins[j] = mins[j - 1];
    freqs[j] = freqs[j - 1];
  }
}
void insert(long long n) {
  for (int i = 0; i < 3; i++) {
    if (n == mins[i]) {
      move(i);
      freqs[i]++;
      return;
    }
    if (n < mins[i]) {
      move(i);
      freqs[i] = 1;
      mins[i] = n;
      return;
    }
  }
}
using namespace std;
int main() {
  mins[0] = 2000000000;
  mins[1] = 2000000000 + 1;
  mins[2] = 2000000000 + 2;
  freqs[0] = 0;
  freqs[1] = 0;
  freqs[2] = 0;
  int amt;
  cin >> amt;
  for (int i = 0; i < amt; i++) {
    long long n;
    cin >> n;
    insert(n);
  }
  if (mins[0] < mins[1] && mins[1] < mins[2]) {
    cout << freqs[0] * freqs[1] * freqs[2];
  }
  if (mins[0] == mins[1] && mins[1] < mins[2]) {
    cout << freqs[0] * (freqs[0] - 1) / 2 * freqs[2];
  }
  if (mins[0] < mins[1] && mins[1] == mins[2]) {
    cout << freqs[1] * (freqs[1] - 1) / 2 * freqs[0];
  }
  if (mins[0] == mins[1] && mins[1] == mins[2]) {
    cout << freqs[0] * (freqs[0] - 1) * (freqs[0] - 2) / 6;
  }
  cout << endl;
}

### Prompt

Your task is to create 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;
long long mins[3];
long long freqs[3];
void move(int i) {
  for (int j = 2; j > i; j--) {
    mins[j] = mins[j - 1];
    freqs[j] = freqs[j - 1];
  }
}
void insert(long long n) {
  for (int i = 0; i < 3; i++) {
    if (n == mins[i]) {
      move(i);
      freqs[i]++;
      return;
    }
    if (n < mins[i]) {
      move(i);
      freqs[i] = 1;
      mins[i] = n;
      return;
    }
  }
}
using namespace std;
int main() {
  mins[0] = 2000000000;
  mins[1] = 2000000000 + 1;
  mins[2] = 2000000000 + 2;
  freqs[0] = 0;
  freqs[1] = 0;
  freqs[2] = 0;
  int amt;
  cin >> amt;
  for (int i = 0; i < amt; i++) {
    long long n;
    cin >> n;
    insert(n);
  }
  if (mins[0] < mins[1] && mins[1] < mins[2]) {
    cout << freqs[0] * freqs[1] * freqs[2];
  }
  if (mins[0] == mins[1] && mins[1] < mins[2]) {
    cout << freqs[0] * (freqs[0] - 1) / 2 * freqs[2];
  }
  if (mins[0] < mins[1] && mins[1] == mins[2]) {
    cout << freqs[1] * (freqs[1] - 1) / 2 * freqs[0];
  }
  if (mins[0] == mins[1] && mins[1] == mins[2]) {
    cout << freqs[0] * (freqs[0] - 1) * (freqs[0] - 2) / 6;
  }
  cout << endl;
}

```