#include <bits/stdc++.h>
using namespace std;
int a[100111];
int pre[100111], suc[100111];
int prec[100111], succ[100111];
int n;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  pre[1] = a[1];
  for (int i = 2; i <= n; i++) {
    pre[i] = min(pre[i - 1], a[i]);
  }
  suc[n] = a[n];
  for (int i = n - 1; i >= 1; i--) {
    suc[i] = min(suc[i + 1], a[i]);
  }
  long double ans = 1e100;
  for (int i = 2; i < n; i++) {
    ans = min(ans, (long double)pre[i - 1] * suc[i + 1] * a[i]);
  }
  for (int i = 1; i <= n; i++) {
    if (i > 1 && pre[i] == pre[i - 1]) {
      prec[i] = prec[i - 1] + (a[i] == pre[i]);
    } else {
      prec[i] = (a[i] == pre[i]);
    }
  }
  for (int i = n; i >= 1; i--) {
    if (i < n && suc[i] == suc[i + 1]) {
      succ[i] = succ[i + 1] + (a[i] == suc[i] ? 1 : 0);
    } else {
      succ[i] = (a[i] == suc[i] ? 1 : 0);
    }
  }
  long long tot = 0;
  for (int i = 2; i < n; i++) {
    if ((long double)pre[i - 1] * suc[i + 1] * a[i] == ans) {
      tot += 1ll * prec[i - 1] * succ[i + 1];
    }
  }
  cout << tot << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 a[100111];
int pre[100111], suc[100111];
int prec[100111], succ[100111];
int n;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  pre[1] = a[1];
  for (int i = 2; i <= n; i++) {
    pre[i] = min(pre[i - 1], a[i]);
  }
  suc[n] = a[n];
  for (int i = n - 1; i >= 1; i--) {
    suc[i] = min(suc[i + 1], a[i]);
  }
  long double ans = 1e100;
  for (int i = 2; i < n; i++) {
    ans = min(ans, (long double)pre[i - 1] * suc[i + 1] * a[i]);
  }
  for (int i = 1; i <= n; i++) {
    if (i > 1 && pre[i] == pre[i - 1]) {
      prec[i] = prec[i - 1] + (a[i] == pre[i]);
    } else {
      prec[i] = (a[i] == pre[i]);
    }
  }
  for (int i = n; i >= 1; i--) {
    if (i < n && suc[i] == suc[i + 1]) {
      succ[i] = succ[i + 1] + (a[i] == suc[i] ? 1 : 0);
    } else {
      succ[i] = (a[i] == suc[i] ? 1 : 0);
    }
  }
  long long tot = 0;
  for (int i = 2; i < n; i++) {
    if ((long double)pre[i - 1] * suc[i + 1] * a[i] == ans) {
      tot += 1ll * prec[i - 1] * succ[i + 1];
    }
  }
  cout << tot << endl;
  return 0;
}

```