#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, arr[100005], a, b, c, idx, cta, ctb, ctc;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  a = b = c = 0;
  cta = ctb = ctc = 0;
  idx = 1;
  sort(arr + 1, arr + 1 + n);
  for (int i = 1; i <= n; i++) {
    if (a == 0) {
      a = arr[i];
    } else if (b == 0 && a != arr[i]) {
      b = arr[i];
    } else if (c == 0 && a != arr[i] && b != arr[i]) {
      c = arr[i];
    }
    if (a == arr[i]) cta++;
    if (b == arr[i]) ctb++;
    if (c == arr[i]) ctc++;
  }
  if (cta >= 3)
    cout << (cta * (cta - 1) * (cta - 2)) / 6 << '\n';
  else if (cta == 2)
    cout << ctb << '\n';
  else {
    if (ctb == 1)
      cout << ctc << '\n';
    else
      cout << (ctb * (ctb - 1)) / 2 << '\n';
  }
}

### 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() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, arr[100005], a, b, c, idx, cta, ctb, ctc;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  a = b = c = 0;
  cta = ctb = ctc = 0;
  idx = 1;
  sort(arr + 1, arr + 1 + n);
  for (int i = 1; i <= n; i++) {
    if (a == 0) {
      a = arr[i];
    } else if (b == 0 && a != arr[i]) {
      b = arr[i];
    } else if (c == 0 && a != arr[i] && b != arr[i]) {
      c = arr[i];
    }
    if (a == arr[i]) cta++;
    if (b == arr[i]) ctb++;
    if (c == arr[i]) ctc++;
  }
  if (cta >= 3)
    cout << (cta * (cta - 1) * (cta - 2)) / 6 << '\n';
  else if (cta == 2)
    cout << ctb << '\n';
  else {
    if (ctb == 1)
      cout << ctc << '\n';
    else
      cout << (ctb * (ctb - 1)) / 2 << '\n';
  }
}

```