#include <bits/stdc++.h>
using namespace std;
long long n, a[100005], b[100005];
struct cell {
  long long kol, val;
};
cell minl[100005], minr[100005];
vector<long long> v1, v2;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b + 1, b + n + 1);
  minl[1].kol = 1;
  minl[1].val = a[1];
  for (int i = 2; i <= n; i++)
    if (a[i] < minl[i - 1].val) {
      minl[i].val = a[i];
      minl[i].kol = 1;
    } else if (a[i] == minl[i - 1].val) {
      minl[i].kol = minl[i - 1].kol + 1;
      minl[i].val = minl[i - 1].val;
    } else {
      minl[i].kol = minl[i - 1].kol;
      minl[i].val = minl[i - 1].val;
    }
  minr[n].kol = 1;
  minr[n].val = a[n];
  for (int i = n - 1; i >= 1; i--)
    if (a[i] < minr[i + 1].val) {
      minr[i].val = a[i];
      minr[i].kol = 1;
    } else if (a[i] == minr[i + 1].val) {
      minr[i].kol = minr[i + 1].kol + 1;
      minr[i].val = minr[i + 1].val;
    } else {
      minr[i].kol = minr[i + 1].kol;
      minr[i].val = minr[i + 1].val;
    }
  v1.push_back(b[1]);
  v1.push_back(b[2]);
  v1.push_back(b[3]);
  long long ans = 0;
  for (int i = 2; i < n; i++) {
    v2.clear();
    v2.push_back(minl[i - 1].val);
    v2.push_back(minr[i + 1].val);
    v2.push_back(a[i]);
    sort(v2.begin(), v2.end());
    if (v1 == v2) ans += minl[i - 1].kol * minr[i + 1].kol;
  }
  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;
long long n, a[100005], b[100005];
struct cell {
  long long kol, val;
};
cell minl[100005], minr[100005];
vector<long long> v1, v2;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b + 1, b + n + 1);
  minl[1].kol = 1;
  minl[1].val = a[1];
  for (int i = 2; i <= n; i++)
    if (a[i] < minl[i - 1].val) {
      minl[i].val = a[i];
      minl[i].kol = 1;
    } else if (a[i] == minl[i - 1].val) {
      minl[i].kol = minl[i - 1].kol + 1;
      minl[i].val = minl[i - 1].val;
    } else {
      minl[i].kol = minl[i - 1].kol;
      minl[i].val = minl[i - 1].val;
    }
  minr[n].kol = 1;
  minr[n].val = a[n];
  for (int i = n - 1; i >= 1; i--)
    if (a[i] < minr[i + 1].val) {
      minr[i].val = a[i];
      minr[i].kol = 1;
    } else if (a[i] == minr[i + 1].val) {
      minr[i].kol = minr[i + 1].kol + 1;
      minr[i].val = minr[i + 1].val;
    } else {
      minr[i].kol = minr[i + 1].kol;
      minr[i].val = minr[i + 1].val;
    }
  v1.push_back(b[1]);
  v1.push_back(b[2]);
  v1.push_back(b[3]);
  long long ans = 0;
  for (int i = 2; i < n; i++) {
    v2.clear();
    v2.push_back(minl[i - 1].val);
    v2.push_back(minr[i + 1].val);
    v2.push_back(a[i]);
    sort(v2.begin(), v2.end());
    if (v1 == v2) ans += minl[i - 1].kol * minr[i + 1].kol;
  }
  cout << ans << endl;
  return 0;
}

```