#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 10;
long long a[N];
map<long long, int> freq1, freq2;
set<long long> st;
long long choose(long long n, long long k) {
  if (k == 0) return 1;
  return (n * choose(n - 1, k - 1)) / k;
}
int main() {
  long long n;
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%I64d", &a[i]);
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    if (i < 3) freq2[a[i]]++;
    freq1[a[i]]++;
  }
  st.insert(a[0]);
  st.insert(a[1]);
  st.insert(a[2]);
  long long ans = 1;
  for (set<long long>::iterator it = st.begin(); it != st.end(); it++)
    ans = ans * choose(freq1[*it], freq2[*it]);
  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 long long N = 1e5 + 10;
long long a[N];
map<long long, int> freq1, freq2;
set<long long> st;
long long choose(long long n, long long k) {
  if (k == 0) return 1;
  return (n * choose(n - 1, k - 1)) / k;
}
int main() {
  long long n;
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%I64d", &a[i]);
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    if (i < 3) freq2[a[i]]++;
    freq1[a[i]]++;
  }
  st.insert(a[0]);
  st.insert(a[1]);
  st.insert(a[2]);
  long long ans = 1;
  for (set<long long>::iterator it = st.begin(); it != st.end(); it++)
    ans = ans * choose(freq1[*it], freq2[*it]);
  cout << ans;
}

```