#include <bits/stdc++.h>
using namespace std;
int n;
const int nm = 1e5 + 10;
int a[nm];
long long A[nm][5];
int x1, x2, x3;
void solve(int x) {
  if (x == 0)
    ++x1;
  else if (x == 1)
    ++x2;
  else if (x == 2)
    ++x3;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  A[1][0] = 1;
  A[1][1] = 1;
  for (int i = 2; i <= nm; i++) {
    A[i][0] = 1;
    for (int j = 1; j <= 3; j++) A[i][j] = A[i - 1][j] + A[i - 1][j - 1];
  }
  while (cin >> n) {
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a, a + n);
    int last = a[0];
    x1 = 0, x2 = 0, x3 = 0;
    for (int i = 0, ct = 0; i < n && ct < 3; i++) {
      if (a[i] == last)
        solve(ct);
      else {
        last = a[i];
        ++ct;
        solve(ct);
      }
    }
    long long ans = 0;
    if (x1 >= 3)
      ans = A[x1][3];
    else if (x1 + x2 >= 3)
      ans = (x1 > 1 ? A[x1][2] * x2 : A[x2][2]);
    else
      ans = x1 * x2 * x3;
    cout << ans << endl;
  }
  return 0;
}

### 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 n;
const int nm = 1e5 + 10;
int a[nm];
long long A[nm][5];
int x1, x2, x3;
void solve(int x) {
  if (x == 0)
    ++x1;
  else if (x == 1)
    ++x2;
  else if (x == 2)
    ++x3;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  A[1][0] = 1;
  A[1][1] = 1;
  for (int i = 2; i <= nm; i++) {
    A[i][0] = 1;
    for (int j = 1; j <= 3; j++) A[i][j] = A[i - 1][j] + A[i - 1][j - 1];
  }
  while (cin >> n) {
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a, a + n);
    int last = a[0];
    x1 = 0, x2 = 0, x3 = 0;
    for (int i = 0, ct = 0; i < n && ct < 3; i++) {
      if (a[i] == last)
        solve(ct);
      else {
        last = a[i];
        ++ct;
        solve(ct);
      }
    }
    long long ans = 0;
    if (x1 >= 3)
      ans = A[x1][3];
    else if (x1 + x2 >= 3)
      ans = (x1 > 1 ? A[x1][2] * x2 : A[x2][2]);
    else
      ans = x1 * x2 * x3;
    cout << ans << endl;
  }
  return 0;
}

```