#include <bits/stdc++.h>
using namespace std;
long long combinatorial(long long x1, long long x2) {
  long long sum = 1, sum2 = 1;
  long long i;
  for (i = x1; i > x1 - x2; i--) sum *= i;
  for (i = 1; i <= x2; i++) sum2 *= i;
  return sum / sum2;
}
int main() {
  long long int i, n, v[100000], count, x[3];
  while (scanf("%lld", &n) != EOF) {
    for (i = 0; i < n; i++) scanf("%lld", &v[i]);
    sort(v, v + n);
    count = 0;
    if (v[0] == v[1] && v[1] == v[2]) {
      for (i = 0; i < n; i++) {
        if (v[i] == v[0]) count++;
      }
      printf("%lld\n", combinatorial(count, 3));
      continue;
    } else if (v[0] == v[1]) {
      for (i = 0; i < n; i++) {
        if (v[i] == v[2]) count++;
      }
      printf("%lld\n", count);
      continue;
    } else if (v[1] == v[2]) {
      for (i = 0; i < n; i++) {
        if (v[i] == v[1]) count++;
      }
      printf("%lld\n", combinatorial(count, 2));
      continue;
    } else if (v[0] != v[1] && v[1] != v[2]) {
      for (i = 0; i < n; i++) {
        if (v[i] == v[2]) count++;
      }
      printf("%lld\n", count);
      continue;
    }
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 combinatorial(long long x1, long long x2) {
  long long sum = 1, sum2 = 1;
  long long i;
  for (i = x1; i > x1 - x2; i--) sum *= i;
  for (i = 1; i <= x2; i++) sum2 *= i;
  return sum / sum2;
}
int main() {
  long long int i, n, v[100000], count, x[3];
  while (scanf("%lld", &n) != EOF) {
    for (i = 0; i < n; i++) scanf("%lld", &v[i]);
    sort(v, v + n);
    count = 0;
    if (v[0] == v[1] && v[1] == v[2]) {
      for (i = 0; i < n; i++) {
        if (v[i] == v[0]) count++;
      }
      printf("%lld\n", combinatorial(count, 3));
      continue;
    } else if (v[0] == v[1]) {
      for (i = 0; i < n; i++) {
        if (v[i] == v[2]) count++;
      }
      printf("%lld\n", count);
      continue;
    } else if (v[1] == v[2]) {
      for (i = 0; i < n; i++) {
        if (v[i] == v[1]) count++;
      }
      printf("%lld\n", combinatorial(count, 2));
      continue;
    } else if (v[0] != v[1] && v[1] != v[2]) {
      for (i = 0; i < n; i++) {
        if (v[i] == v[2]) count++;
      }
      printf("%lld\n", count);
      continue;
    }
  }
  return 0;
}

```