#include <bits/stdc++.h>
using namespace std;
long long n, sb, a[4], data1[100001], num[100001];
void merge(int l, int r) {
  int mid = (l + r) / 2;
  int i = l, j = mid + 1, k = 0;
  while (i <= mid && j <= r) {
    if (data1[i] < data1[j])
      num[k++] = data1[i++];
    else
      num[k++] = data1[j++];
  }
  while (i <= mid) num[k++] = data1[i++];
  while (j <= r) num[k++] = data1[j++];
  for (int p = 0; p < k; p++) data1[p + l] = num[p];
}
void sort(int l, int r) {
  if (l >= r) return;
  int mid = (l + r) / 2;
  sort(l, mid);
  sort(mid + 1, r);
  merge(l, r);
}
int main() {
  cin >> n;
  for (int j = 0; j < n; j++) cin >> data1[j];
  sort(0, n - 1);
  for (int i = 1; i <= 3; i++) {
    a[i] = 1;
    while (data1[sb] == data1[sb + 1] && sb < n - 1) {
      a[i]++;
      sb++;
    }
    sb++;
  }
  if (data1[0] == data1[1]) {
    if (data1[1] == data1[2])
      cout << a[1] * (a[1] - 1) * (a[1] - 2) / 6;
    else
      cout << a[2];
  } else {
    if (data1[1] == data1[2])
      cout << a[2] * (a[2] - 1) / 2;
    else
      cout << a[3];
  }
}

### 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;
long long n, sb, a[4], data1[100001], num[100001];
void merge(int l, int r) {
  int mid = (l + r) / 2;
  int i = l, j = mid + 1, k = 0;
  while (i <= mid && j <= r) {
    if (data1[i] < data1[j])
      num[k++] = data1[i++];
    else
      num[k++] = data1[j++];
  }
  while (i <= mid) num[k++] = data1[i++];
  while (j <= r) num[k++] = data1[j++];
  for (int p = 0; p < k; p++) data1[p + l] = num[p];
}
void sort(int l, int r) {
  if (l >= r) return;
  int mid = (l + r) / 2;
  sort(l, mid);
  sort(mid + 1, r);
  merge(l, r);
}
int main() {
  cin >> n;
  for (int j = 0; j < n; j++) cin >> data1[j];
  sort(0, n - 1);
  for (int i = 1; i <= 3; i++) {
    a[i] = 1;
    while (data1[sb] == data1[sb + 1] && sb < n - 1) {
      a[i]++;
      sb++;
    }
    sb++;
  }
  if (data1[0] == data1[1]) {
    if (data1[1] == data1[2])
      cout << a[1] * (a[1] - 1) * (a[1] - 2) / 6;
    else
      cout << a[2];
  } else {
    if (data1[1] == data1[2])
      cout << a[2] * (a[2] - 1) / 2;
    else
      cout << a[3];
  }
}

```