#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7;
int n, a[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  sort(a + 1, a + 1 + n);
  if (a[1] == a[2] && a[2] == a[3]) {
    int cnt = 0;
    for (int i = 1; i <= n; ++i)
      if (a[i] != a[1])
        break;
      else
        cnt++;
    printf("%I64d\n", 1ll * cnt * (cnt - 1) * (cnt - 2) / 6);
  } else if (a[1] == a[2] && a[2] != a[3]) {
    int cnt1 = 0, cnt2 = 0;
    for (int i = 1; i <= n; ++i) {
      if (a[i] == a[1]) cnt1++;
      if (a[i] == a[3]) cnt2++;
    }
    printf("%I64d\n", 1ll * cnt1 * (cnt1 - 1) / 2 * cnt2);
  } else if (a[1] != a[2] && a[2] == a[3]) {
    int cnt1 = 0, cnt2 = 0;
    for (int i = 1; i <= n; ++i) {
      if (a[i] == a[2]) cnt1++;
      if (a[i] == a[1]) cnt2++;
    }
    printf("%I64d\n", 1ll * cnt1 * (cnt1 - 1) / 2 * cnt2);
  } else {
    int cnt1 = 0, cnt2 = 0, cnt3 = 0;
    for (int i = 1; i <= n; ++i) {
      if (a[i] == a[1]) cnt1++;
      if (a[i] == a[2]) cnt2++;
      if (a[i] == a[3]) cnt3++;
    }
    printf("%I64d\n", 1ll * cnt1 * cnt2 * cnt3);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
const int N = 1e5 + 7;
int n, a[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  sort(a + 1, a + 1 + n);
  if (a[1] == a[2] && a[2] == a[3]) {
    int cnt = 0;
    for (int i = 1; i <= n; ++i)
      if (a[i] != a[1])
        break;
      else
        cnt++;
    printf("%I64d\n", 1ll * cnt * (cnt - 1) * (cnt - 2) / 6);
  } else if (a[1] == a[2] && a[2] != a[3]) {
    int cnt1 = 0, cnt2 = 0;
    for (int i = 1; i <= n; ++i) {
      if (a[i] == a[1]) cnt1++;
      if (a[i] == a[3]) cnt2++;
    }
    printf("%I64d\n", 1ll * cnt1 * (cnt1 - 1) / 2 * cnt2);
  } else if (a[1] != a[2] && a[2] == a[3]) {
    int cnt1 = 0, cnt2 = 0;
    for (int i = 1; i <= n; ++i) {
      if (a[i] == a[2]) cnt1++;
      if (a[i] == a[1]) cnt2++;
    }
    printf("%I64d\n", 1ll * cnt1 * (cnt1 - 1) / 2 * cnt2);
  } else {
    int cnt1 = 0, cnt2 = 0, cnt3 = 0;
    for (int i = 1; i <= n; ++i) {
      if (a[i] == a[1]) cnt1++;
      if (a[i] == a[2]) cnt2++;
      if (a[i] == a[3]) cnt3++;
    }
    printf("%I64d\n", 1ll * cnt1 * cnt2 * cnt3);
  }
  return 0;
}

```