Makes And The Product

After returning from the army Makes received a gift — an array a consisting of npositive 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.

Example

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.

有三种情况:

1.最小数字的个数n1>=3,则只需计算


2.n1<3,第二小的数字个数n2>=3-n1,则只需计算


3.n1,n2<3,第三小的数字个数n3>=3-n2-n1,则只需计算


注意结果会超过int

#include<iostream>
#include<queue>
#include<map>
using namespace std;
struct list
{
int num,amount;
list() { }
list(int a,int b):num(a),amount(b) { }
friend bool operator < (struct list c,struct list d)
{
return c.num>d.num;
}
};
long long C(int a,int b)
{
if(b>a-b)
b=a-b;
if(b==0)
return 1;
long long up=1,down=1;
int i=b;
while(i--)
up*=a--;
for(i=1;i<=b;i++)
down*=i;
return up/down;
}
int main()
{
int n;
while(cin >> n)
{
map<int,int> mp;
priority_queue<list> q;
while(n--)
{
int a;
cin >> a;
mp[a]++;
}
map<int,int>::iterator it;
for(it=mp.begin();it!=mp.end();it++)
q.push(list(it->first,it->second));
if(q.top().amount>=3)
{
cout << C(q.top().amount,3) << endl;
continue;
}
else
{
int num=q.top().amount;
q.pop();
if(q.top().amount>=3-num)
{
cout << C(q.top().amount,3-num) << endl;
continue;
}
else
{
num+=q.top().amount;
q.pop();
cout << C(q.top().amount,3-num) << endl;
}
}
}
return 0;
}