#include <bits/stdc++.h>
using namespace std;
long long a[1000010], d[1000010];
long long b[100010], c[1010][1010];
long long mn = 1e18, mx = -1e18, ans, cnt, sm = 1, n, m, cur;
bool ok, okk, u[1000010], uu[310][310], uk[1000010];
char ck[510][510], ch[100010];
int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
vector<long long> g[900010];
int main() {
  std::ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  long long x = a[0], y = a[1], z = a[2];
  if (x == y && x == z) {
    for (int i = 0; i < n; i++) {
      if (a[i] == x) {
        cnt++;
      }
    }
    long long k = cnt * (cnt - 1) * (cnt - 2) / 6;
    cout << k;
    return 0;
  }
  if (x == y) {
    for (int i = 0; i < n; i++) {
      if (a[i] == z) {
        cnt++;
      }
    }
    cout << cnt << endl;
    return 0;
  }
  if (y == z) {
    for (int i = 0; i < n; i++) {
      if (a[i] == y) {
        cnt++;
      }
    }
    cout << ((cnt - 1) * cnt) / 2;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (a[i] == z) {
      cnt++;
    }
  }
  cout << cnt;
  return 0;
}

### Prompt

Please formulate 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 a[1000010], d[1000010];
long long b[100010], c[1010][1010];
long long mn = 1e18, mx = -1e18, ans, cnt, sm = 1, n, m, cur;
bool ok, okk, u[1000010], uu[310][310], uk[1000010];
char ck[510][510], ch[100010];
int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
vector<long long> g[900010];
int main() {
  std::ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  long long x = a[0], y = a[1], z = a[2];
  if (x == y && x == z) {
    for (int i = 0; i < n; i++) {
      if (a[i] == x) {
        cnt++;
      }
    }
    long long k = cnt * (cnt - 1) * (cnt - 2) / 6;
    cout << k;
    return 0;
  }
  if (x == y) {
    for (int i = 0; i < n; i++) {
      if (a[i] == z) {
        cnt++;
      }
    }
    cout << cnt << endl;
    return 0;
  }
  if (y == z) {
    for (int i = 0; i < n; i++) {
      if (a[i] == y) {
        cnt++;
      }
    }
    cout << ((cnt - 1) * cnt) / 2;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (a[i] == z) {
      cnt++;
    }
  }
  cout << cnt;
  return 0;
}

```