#include <bits/stdc++.h>
using namespace std;
const int Mod = 1000000007;
const int N = 100005;
int n, arr[N];
long long dp[N][3];
long long calc(int i, int ind) {
  if (ind == 3) return 1;
  if (i == n) return 0;
  if (arr[i] != arr[ind]) return 0;
  long long &r = dp[i][ind];
  if (r != -1) return r;
  r = calc(i + 1, ind) + calc(i + 1, ind + 1);
  return r;
}
int main() {
  memset(dp, -1, sizeof dp);
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  sort(arr, arr + n);
  printf("%lld\n", calc(0, 0));
  return 0;
}

### Prompt

Create a solution in CPP for 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 = 1000000007;
const int N = 100005;
int n, arr[N];
long long dp[N][3];
long long calc(int i, int ind) {
  if (ind == 3) return 1;
  if (i == n) return 0;
  if (arr[i] != arr[ind]) return 0;
  long long &r = dp[i][ind];
  if (r != -1) return r;
  r = calc(i + 1, ind) + calc(i + 1, ind + 1);
  return r;
}
int main() {
  memset(dp, -1, sizeof dp);
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  sort(arr, arr + n);
  printf("%lld\n", calc(0, 0));
  return 0;
}

```