#include <bits/stdc++.h>
int main() {
  long long a[100003], i, j, n, b, c, d, sum1, sum2;
  while (scanf("%lld", &n) != EOF) {
    for (i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
    }
    for (i = 0; i < 3; i++) {
      for (j = i + 1; j < n; j++) {
        if (a[i] > a[j]) {
          b = a[i];
          a[i] = a[j];
          a[j] = b;
        }
      }
    }
    b = a[2];
    c = 1;
    d = 0;
    for (i = 0; i < n; i++) {
      if (i == 0 && a[i] == b) c++;
      if (i == 1 && a[i] == b) c++;
      if (a[i] == b) d++;
    }
    sum1 = sum2 = 1;
    while (c > 0) {
      sum1 *= c;
      sum2 *= d;
      c--;
      d--;
    }
    printf("%lld\n", sum2 / sum1);
  }
  return 0;
}

### 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>
int main() {
  long long a[100003], i, j, n, b, c, d, sum1, sum2;
  while (scanf("%lld", &n) != EOF) {
    for (i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
    }
    for (i = 0; i < 3; i++) {
      for (j = i + 1; j < n; j++) {
        if (a[i] > a[j]) {
          b = a[i];
          a[i] = a[j];
          a[j] = b;
        }
      }
    }
    b = a[2];
    c = 1;
    d = 0;
    for (i = 0; i < n; i++) {
      if (i == 0 && a[i] == b) c++;
      if (i == 1 && a[i] == b) c++;
      if (a[i] == b) d++;
    }
    sum1 = sum2 = 1;
    while (c > 0) {
      sum1 *= c;
      sum2 *= d;
      c--;
      d--;
    }
    printf("%lld\n", sum2 / sum1);
  }
  return 0;
}

```