#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-9;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long i, j, k, l, p, q, r, x, y, z, n, m, c, t, h;
  cin >> n;
  long long arr[n];
  map<long long, long long> mp;
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    mp[arr[i]]++;
  }
  map<long long, long long>::iterator it = mp.begin();
  long long mx = 0;
  c = 0;
  vector<long long> v;
  for (i = 1; it != mp.end() && i <= 3; it++, i++) {
    c += mp[it->first];
    v.push_back(mp[it->first]);
    if (c >= 3) break;
  }
  if (i == 1) {
    k = 1;
    for (j = mp[it->first] - 2; j <= mp[it->first]; j++) k *= j;
    cout << k / 6 << endl;
  } else if (i == 2) {
    if (c == 3)
      cout << 1 << endl;
    else if (min(v[0], v[1]) == 1) {
      mx = max(v[0], v[1]);
      mx--;
      cout << (mx * (mx + 1)) / 2 << endl;
    } else
      cout << max(v[0], v[1]);
  } else {
    mx = max(max(v[0], v[1]), v[2]);
    cout << mx << endl;
  }
}

### Prompt

Generate 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 double PI = acos(-1);
const double eps = 1e-9;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long i, j, k, l, p, q, r, x, y, z, n, m, c, t, h;
  cin >> n;
  long long arr[n];
  map<long long, long long> mp;
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    mp[arr[i]]++;
  }
  map<long long, long long>::iterator it = mp.begin();
  long long mx = 0;
  c = 0;
  vector<long long> v;
  for (i = 1; it != mp.end() && i <= 3; it++, i++) {
    c += mp[it->first];
    v.push_back(mp[it->first]);
    if (c >= 3) break;
  }
  if (i == 1) {
    k = 1;
    for (j = mp[it->first] - 2; j <= mp[it->first]; j++) k *= j;
    cout << k / 6 << endl;
  } else if (i == 2) {
    if (c == 3)
      cout << 1 << endl;
    else if (min(v[0], v[1]) == 1) {
      mx = max(v[0], v[1]);
      mx--;
      cout << (mx * (mx + 1)) / 2 << endl;
    } else
      cout << max(v[0], v[1]);
  } else {
    mx = max(max(v[0], v[1]), v[2]);
    cout << mx << endl;
  }
}

```