#include <bits/stdc++.h>
using namespace std;
long long a[1000005];
unordered_map<long long, long long> mp[100005];
long long ncr(long long n, long long r) {
  if (r > n) return 0;
  if (!r) return 1;
  if (mp[n].find(r) != mp[n].end()) return mp[n][r];
  return mp[n][r] = ncr(n - 1, r - 1) + ncr(n - 1, r);
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n, m, k, ans = 1, sum = 0, x, y, p, tot = 0;
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; cin >> a[i++])
    ;
  sort(a, a + n);
  for (sum = 1; sum < n && a[sum] == a[0]; ++sum)
    ;
  v.push_back(sum);
  tot += sum;
  if (tot > 2) return cout << ncr(v[0], 3), 0;
  ans = sum;
  for (sum = ans + 1; sum < n && a[sum] == a[ans]; ++sum)
    ;
  v.push_back(sum - ans);
  tot += sum - ans;
  if (tot > 2) return cout << ncr(v[1], 3 - v[0]), 0;
  ans = sum;
  for (sum = ans + 1; sum < n && a[sum] == a[ans]; ++sum)
    ;
  v.push_back(sum - ans);
  cout << ncr(v[2], 3 - v[0] - v[1]);
}

### Prompt

Please create a solution in Cpp 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 a[1000005];
unordered_map<long long, long long> mp[100005];
long long ncr(long long n, long long r) {
  if (r > n) return 0;
  if (!r) return 1;
  if (mp[n].find(r) != mp[n].end()) return mp[n][r];
  return mp[n][r] = ncr(n - 1, r - 1) + ncr(n - 1, r);
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n, m, k, ans = 1, sum = 0, x, y, p, tot = 0;
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; cin >> a[i++])
    ;
  sort(a, a + n);
  for (sum = 1; sum < n && a[sum] == a[0]; ++sum)
    ;
  v.push_back(sum);
  tot += sum;
  if (tot > 2) return cout << ncr(v[0], 3), 0;
  ans = sum;
  for (sum = ans + 1; sum < n && a[sum] == a[ans]; ++sum)
    ;
  v.push_back(sum - ans);
  tot += sum - ans;
  if (tot > 2) return cout << ncr(v[1], 3 - v[0]), 0;
  ans = sum;
  for (sum = ans + 1; sum < n && a[sum] == a[ans]; ++sum)
    ;
  v.push_back(sum - ans);
  cout << ncr(v[2], 3 - v[0] - v[1]);
}

```