#include <bits/stdc++.h>
using namespace std;
vector<long long int> v[100005];
long long int dp[100005] = {0};
bool vis[100005] = {false};
long long int mx = -1e13;
long long int o;
void swap(int &a, int &b) {
  auto tm = a;
  a = b;
  b = tm;
}
long long int power(long long int x, long long int y) {
  long long int sum = 1;
  for (long long int i = 0; i < y; i++) sum *= x;
  return sum;
}
long long int fact(long long int n) {
  long long int res = 1;
  for (long long int i = 2; i <= n; i++) res = res * i;
  return res;
}
long long int nCr(long long int n, long long int r) {
  return fact(n) / (fact(r) * fact(n - r));
}
void jai_mata_di() {
  long long int q;
  q = 1;
  long long int u = 0;
  while (q--) {
    long long int n, i, j, s, x, y, sum = 0, mn = 1e13;
    long long int x1, y1, x2, y2;
    map<long long int, long long int> m;
    cin >> n;
    long long int a[n];
    for (i = 0; i < n; i++) cin >> a[i];
    sort(a, a + n);
    long long int a1, b1, c = 1;
    for (i = 3; i < n; i++) {
      if (a[i] == a[i - 1])
        c++;
      else
        break;
    }
    if (a[0] == a[1] && a[1] == a[2]) {
      cout << ((c + 2) * (c + 1) * (c)) / 6;
    } else if (a[2] == a[1])
      cout << ((c + 1) * c) / 2;
    else
      cout << c;
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int tt = 1.0 * clock();
  int t;
  t = 1;
  for (int y = 0; y < t; y++) {
    jai_mata_di();
    cout << endl;
    return 0;
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
vector<long long int> v[100005];
long long int dp[100005] = {0};
bool vis[100005] = {false};
long long int mx = -1e13;
long long int o;
void swap(int &a, int &b) {
  auto tm = a;
  a = b;
  b = tm;
}
long long int power(long long int x, long long int y) {
  long long int sum = 1;
  for (long long int i = 0; i < y; i++) sum *= x;
  return sum;
}
long long int fact(long long int n) {
  long long int res = 1;
  for (long long int i = 2; i <= n; i++) res = res * i;
  return res;
}
long long int nCr(long long int n, long long int r) {
  return fact(n) / (fact(r) * fact(n - r));
}
void jai_mata_di() {
  long long int q;
  q = 1;
  long long int u = 0;
  while (q--) {
    long long int n, i, j, s, x, y, sum = 0, mn = 1e13;
    long long int x1, y1, x2, y2;
    map<long long int, long long int> m;
    cin >> n;
    long long int a[n];
    for (i = 0; i < n; i++) cin >> a[i];
    sort(a, a + n);
    long long int a1, b1, c = 1;
    for (i = 3; i < n; i++) {
      if (a[i] == a[i - 1])
        c++;
      else
        break;
    }
    if (a[0] == a[1] && a[1] == a[2]) {
      cout << ((c + 2) * (c + 1) * (c)) / 6;
    } else if (a[2] == a[1])
      cout << ((c + 1) * c) / 2;
    else
      cout << c;
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int tt = 1.0 * clock();
  int t;
  t = 1;
  for (int y = 0; y < t; y++) {
    jai_mata_di();
    cout << endl;
    return 0;
  }
}

```