#include <bits/stdc++.h>
using namespace std;
map<long long int, long long int> mp;
long long int dp[100005][5] = {-1};
long long int NCR(long long int a, long long int b) {
  if (b > a) return 0;
  if (b * 2 > a) b = a - b;
  if (b == 0) return 1;
  long long int result = a;
  for (int i = 2; i <= b; ++i) {
    result *= (a - i + 1);
    result /= i;
  }
  return result;
};
int main() {
  long long int n, a[100005], mx = 0, s = 0, x, y;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]]++;
  }
  std::map<long long int, long long int>::iterator it;
  int i = 0;
  for (it = mp.begin(); i < 3; i++, it++) {
    s = s + it->second;
    x = it->first;
    if (s >= 3) break;
  }
  if (i == 0) {
    y = NCR(mp[x], 3);
    cout << y << endl;
  } else if (s == 3)
    cout << "1" << endl;
  else if (i == 1) {
    y = 3 - (s - mp[x]);
    y = NCR(mp[x], y);
    cout << y << endl;
  } else if (i == 2) {
    y = NCR(mp[x], 1);
    cout << y << endl;
  }
}

### Prompt

Your task is to create 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;
map<long long int, long long int> mp;
long long int dp[100005][5] = {-1};
long long int NCR(long long int a, long long int b) {
  if (b > a) return 0;
  if (b * 2 > a) b = a - b;
  if (b == 0) return 1;
  long long int result = a;
  for (int i = 2; i <= b; ++i) {
    result *= (a - i + 1);
    result /= i;
  }
  return result;
};
int main() {
  long long int n, a[100005], mx = 0, s = 0, x, y;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]]++;
  }
  std::map<long long int, long long int>::iterator it;
  int i = 0;
  for (it = mp.begin(); i < 3; i++, it++) {
    s = s + it->second;
    x = it->first;
    if (s >= 3) break;
  }
  if (i == 0) {
    y = NCR(mp[x], 3);
    cout << y << endl;
  } else if (s == 3)
    cout << "1" << endl;
  else if (i == 1) {
    y = 3 - (s - mp[x]);
    y = NCR(mp[x], y);
    cout << y << endl;
  } else if (i == 2) {
    y = NCR(mp[x], 1);
    cout << y << endl;
  }
}

```