#include <bits/stdc++.h>
using namespace std;
long long int C(long long int n, long long int r) {
  if (r == 3) {
    long long int ans = 1;
    ans = ans * n;
    ans = ans * (n - 1);
    ans = ans * (n - 2);
    ans = ans / 6;
    return (ans);
  } else {
    long long int ans = 1;
    ans = ans * n;
    ans = ans * (n - 1);
    ans = ans / 2;
    return (ans);
  }
}
int main() {
  int n;
  map<int, int> m;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int input;
    cin >> input;
    if (m.find(input) != m.end())
      m[input]++;
    else
      m[input] = 1;
  }
  int n1, n2, n3;
  n1 = m.begin()->second;
  m.erase(m.begin());
  if (!m.empty()) {
    n2 = m.begin()->second;
    m.erase(m.begin());
  }
  if (!m.empty()) n3 = m.begin()->second;
  if (n1 >= 3) {
    cout << C((long long int)n1, 3);
    return (0);
  }
  if (n1 == 2) {
    cout << n2;
    return (0);
  }
  if (n1 == 1 && n2 != 1) {
    cout << C(n2, 2);
    return (0);
  } else {
    cout << n3;
    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 int C(long long int n, long long int r) {
  if (r == 3) {
    long long int ans = 1;
    ans = ans * n;
    ans = ans * (n - 1);
    ans = ans * (n - 2);
    ans = ans / 6;
    return (ans);
  } else {
    long long int ans = 1;
    ans = ans * n;
    ans = ans * (n - 1);
    ans = ans / 2;
    return (ans);
  }
}
int main() {
  int n;
  map<int, int> m;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int input;
    cin >> input;
    if (m.find(input) != m.end())
      m[input]++;
    else
      m[input] = 1;
  }
  int n1, n2, n3;
  n1 = m.begin()->second;
  m.erase(m.begin());
  if (!m.empty()) {
    n2 = m.begin()->second;
    m.erase(m.begin());
  }
  if (!m.empty()) n3 = m.begin()->second;
  if (n1 >= 3) {
    cout << C((long long int)n1, 3);
    return (0);
  }
  if (n1 == 2) {
    cout << n2;
    return (0);
  }
  if (n1 == 1 && n2 != 1) {
    cout << C(n2, 2);
    return (0);
  } else {
    cout << n3;
    return (0);
  }
}

```