#include <bits/stdc++.h>
using namespace std;
long long n, pos, t, a[100001], ans, c[3];
void solve(long long ind) {
  t = a[pos++];
  c[ind] = 1;
  while (pos < n && a[pos] == t) ++c[ind], ++pos;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  sort(a, a + n);
  pos = 0;
  for (int i = 0; i < 3; ++i) {
    if (pos < n) solve(i);
  }
  if (c[0] >= 3)
    ans = (c[0] * (c[0] - 1) * (c[0] - 2)) / 6;
  else if (c[0] + c[1] >= 3) {
    if (c[0] > 1)
      ans = c[1];
    else
      ans = (c[1] * (c[1] - 1)) / 2;
  } else
    ans = c[2];
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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, pos, t, a[100001], ans, c[3];
void solve(long long ind) {
  t = a[pos++];
  c[ind] = 1;
  while (pos < n && a[pos] == t) ++c[ind], ++pos;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  sort(a, a + n);
  pos = 0;
  for (int i = 0; i < 3; ++i) {
    if (pos < n) solve(i);
  }
  if (c[0] >= 3)
    ans = (c[0] * (c[0] - 1) * (c[0] - 2)) / 6;
  else if (c[0] + c[1] >= 3) {
    if (c[0] > 1)
      ans = c[1];
    else
      ans = (c[1] * (c[1] - 1)) / 2;
  } else
    ans = c[2];
  cout << ans << endl;
  return 0;
}

```