#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int a[100010];
long long vis[3];
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    sort(a + 1, a + n + 1);
    memset(vis, 0, sizeof(vis));
    int cnt = 0;
    vis[0] = 1;
    for (int i = 2; i <= 3; i++)
      if (a[i] == a[i - 1])
        vis[cnt]++;
      else
        vis[++cnt]++;
    for (int i = 4; i <= n; i++)
      if (a[i] == a[i - 1])
        vis[cnt]++;
      else
        break;
    long long ans = 0;
    if (cnt == 0)
      ans = vis[0] * (vis[0] - 1) * (vis[0] - 2) / 6;
    else if (cnt == 1) {
      if (a[2] == a[1])
        ans = vis[1];
      else
        ans = vis[1] * (vis[1] - 1) / 2;
    } else
      ans = vis[2];
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
const int mod = 1e9 + 7;
int a[100010];
long long vis[3];
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    sort(a + 1, a + n + 1);
    memset(vis, 0, sizeof(vis));
    int cnt = 0;
    vis[0] = 1;
    for (int i = 2; i <= 3; i++)
      if (a[i] == a[i - 1])
        vis[cnt]++;
      else
        vis[++cnt]++;
    for (int i = 4; i <= n; i++)
      if (a[i] == a[i - 1])
        vis[cnt]++;
      else
        break;
    long long ans = 0;
    if (cnt == 0)
      ans = vis[0] * (vis[0] - 1) * (vis[0] - 2) / 6;
    else if (cnt == 1) {
      if (a[2] == a[1])
        ans = vis[1];
      else
        ans = vis[1] * (vis[1] - 1) / 2;
    } else
      ans = vis[2];
    printf("%I64d\n", ans);
  }
  return 0;
}

```