#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  int a[100000];
  cin >> a[0] >> a[1] >> a[2];
  long long m1 = a[0], m2 = a[1], m3 = a[2];
  if (m3 < m2) {
    long long t = m2;
    m2 = m3;
    m3 = t;
  }
  if (m2 < m1) {
    long long t = m1;
    m1 = m2;
    m2 = t;
  }
  if (m3 < m2) {
    long long t = m2;
    m2 = m3;
    m3 = t;
  }
  if (m1 == m2 && m2 == m3) {
    m2 = 10000000000;
    m3 = 10000000000;
  } else if (m1 == m2) {
    m2 = m3;
    m3 = 10000000000;
  } else if (m2 == m3)
    m3 = 10000000000;
  for (int i = 3; i < n; i++) {
    cin >> a[i];
    if (a[i] < m3 && a[i] != m2 && a[i] != m1) {
      m3 = a[i];
      if (m3 < m2) {
        int t = m2;
        m2 = m3;
        m3 = t;
      }
      if (m2 < m1) {
        int t = m1;
        m1 = m2;
        m2 = t;
      }
      if (m3 < m2) {
        int t = m2;
        m2 = m3;
        m3 = t;
      }
    }
  }
  long long c1 = 0, c2 = 0, c3 = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == m1) c1++;
    if (a[i] == m2) c2++;
    if (a[i] == m3) c3++;
  }
  long long ans = 1;
  if (c1 > 2) ans = (c1 * (c1 - 1) * (c1 - 2)) / 6;
  if (c1 == 2) {
    ans = c2;
  }
  if (c1 == 1 && c2 > 1) ans = (c2 * (c2 - 1)) / 2;
  if (c1 == 1 && c2 == 1) ans = c3;
  cout << ans;
  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 main() {
  int n;
  cin >> n;
  int a[100000];
  cin >> a[0] >> a[1] >> a[2];
  long long m1 = a[0], m2 = a[1], m3 = a[2];
  if (m3 < m2) {
    long long t = m2;
    m2 = m3;
    m3 = t;
  }
  if (m2 < m1) {
    long long t = m1;
    m1 = m2;
    m2 = t;
  }
  if (m3 < m2) {
    long long t = m2;
    m2 = m3;
    m3 = t;
  }
  if (m1 == m2 && m2 == m3) {
    m2 = 10000000000;
    m3 = 10000000000;
  } else if (m1 == m2) {
    m2 = m3;
    m3 = 10000000000;
  } else if (m2 == m3)
    m3 = 10000000000;
  for (int i = 3; i < n; i++) {
    cin >> a[i];
    if (a[i] < m3 && a[i] != m2 && a[i] != m1) {
      m3 = a[i];
      if (m3 < m2) {
        int t = m2;
        m2 = m3;
        m3 = t;
      }
      if (m2 < m1) {
        int t = m1;
        m1 = m2;
        m2 = t;
      }
      if (m3 < m2) {
        int t = m2;
        m2 = m3;
        m3 = t;
      }
    }
  }
  long long c1 = 0, c2 = 0, c3 = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == m1) c1++;
    if (a[i] == m2) c2++;
    if (a[i] == m3) c3++;
  }
  long long ans = 1;
  if (c1 > 2) ans = (c1 * (c1 - 1) * (c1 - 2)) / 6;
  if (c1 == 2) {
    ans = c2;
  }
  if (c1 == 1 && c2 > 1) ans = (c2 * (c2 - 1)) / 2;
  if (c1 == 1 && c2 == 1) ans = c3;
  cout << ans;
  return 0;
}

```