#include <bits/stdc++.h>
using namespace std;
long long arr[3000000];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(0);
  srand(time(NULL));
  long long n, num = 0, numa = 0, numb = 0, numc = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + n);
  long long a = arr[0], b = arr[1], c = arr[2];
  for (int i = 0; i < n; i++) {
    if (arr[i] == a) numa++;
    if (arr[i] == b) numb++;
    if (arr[i] == c) numc++;
  }
  if (a == b && a == c) {
    long long ans = numa * (numa - 1) * (numa - 2);
    cout << ans / 6;
    return 0;
  } else if ((a == b && a != c) || (a != b && a != c && b != c)) {
    long long ans = numc;
    cout << ans;
    return 0;
  } else if (a != b && b == c) {
    long long ans = numb * (numb - 1);
    cout << ans / 2;
    return 0;
  }
}

### Prompt

In CPP, your task is to solve 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 arr[3000000];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(0);
  srand(time(NULL));
  long long n, num = 0, numa = 0, numb = 0, numc = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + n);
  long long a = arr[0], b = arr[1], c = arr[2];
  for (int i = 0; i < n; i++) {
    if (arr[i] == a) numa++;
    if (arr[i] == b) numb++;
    if (arr[i] == c) numc++;
  }
  if (a == b && a == c) {
    long long ans = numa * (numa - 1) * (numa - 2);
    cout << ans / 6;
    return 0;
  } else if ((a == b && a != c) || (a != b && a != c && b != c)) {
    long long ans = numc;
    cout << ans;
    return 0;
  } else if (a != b && b == c) {
    long long ans = numb * (numb - 1);
    cout << ans / 2;
    return 0;
  }
}

```