#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int mex = 250002;
long long int power(long long int a, long long int b) {
  long long int ans = 1, f = a;
  while (b) {
    if (b & 1) ans = (ans * f) % mod;
    b = b >> 1;
    f = (f * f) % mod;
  }
  return ans % mod;
}
int sl(int a[2][3]) {
  vector<long long int> v;
  for (int i = 0; i < 3; i++) {
    for (int j = i + 1; j < 3; j++) {
      long long int y = (a[0][j] - a[0][i]) * (a[0][j] - a[0][i]) +
                        (a[1][j] - a[1][i]) * (a[1][j] - a[1][i]);
      v.push_back(y);
    }
  }
  sort(v.begin(), v.end());
  if (v[0] == 0 || v[1] == 0 || v[2] == 0) return 0;
  if (v[2] == (v[0] + v[1]))
    return 1;
  else
    return 0;
}
int main() {
  int n;
  cin >> n;
  long long int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  long long int ans = 0;
  map<int, int> mp;
  for (int i = 0; i < n; i++) {
    mp[a[i]]++;
    if (mp.size() == 4) break;
  }
  auto it = mp.begin();
  if (it->second >= 3) {
    long long int u = it->second;
    u = (u * (u - 1) * (u - 2)) / 6;
    ans = u;
  } else if (it->second >= 2) {
    it++;
    long long int u = it->second;
    ans = u;
  } else {
    it++;
    if (it->second == 1) {
      it++;
      ans = it->second;
    } else {
      long long int u = it->second;
      u = (u * (u - 1)) / 2;
      ans = u;
    }
  }
  cout << ans;
}

### 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;
const int mod = 998244353;
const int mex = 250002;
long long int power(long long int a, long long int b) {
  long long int ans = 1, f = a;
  while (b) {
    if (b & 1) ans = (ans * f) % mod;
    b = b >> 1;
    f = (f * f) % mod;
  }
  return ans % mod;
}
int sl(int a[2][3]) {
  vector<long long int> v;
  for (int i = 0; i < 3; i++) {
    for (int j = i + 1; j < 3; j++) {
      long long int y = (a[0][j] - a[0][i]) * (a[0][j] - a[0][i]) +
                        (a[1][j] - a[1][i]) * (a[1][j] - a[1][i]);
      v.push_back(y);
    }
  }
  sort(v.begin(), v.end());
  if (v[0] == 0 || v[1] == 0 || v[2] == 0) return 0;
  if (v[2] == (v[0] + v[1]))
    return 1;
  else
    return 0;
}
int main() {
  int n;
  cin >> n;
  long long int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  long long int ans = 0;
  map<int, int> mp;
  for (int i = 0; i < n; i++) {
    mp[a[i]]++;
    if (mp.size() == 4) break;
  }
  auto it = mp.begin();
  if (it->second >= 3) {
    long long int u = it->second;
    u = (u * (u - 1) * (u - 2)) / 6;
    ans = u;
  } else if (it->second >= 2) {
    it++;
    long long int u = it->second;
    ans = u;
  } else {
    it++;
    if (it->second == 1) {
      it++;
      ans = it->second;
    } else {
      long long int u = it->second;
      u = (u * (u - 1)) / 2;
      ans = u;
    }
  }
  cout << ans;
}

```