Educational Codeforces Round 23 B. Makes And The Product

B. Makes And The Product

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

题意：

给一串数字，在其中选三个数字，使得这三个数字的乘积最小，问有多少个这样的组合

思路：

直接排序乘积最小那么这三个数字必定是最小的三个，又因为数字可能相同，会产成不同结果的情况一共有三种：

1.三个数字都相同 。

2.第一个和第二个不同，第二个与第三个相同。

3.第二个和第三个不同。

实现代码：

#include<iostream>
#include<algorithm>
#include<map>
using namespace std;
#define ll long long
map<ll,ll>mp;
int main()
{
    ll m,i,a[100009];
    cin>>m;
    for(i=0;i<m;i++){
        cin>>a[i];
        mp[a[i]]++;}
    sort(a,a+m);
    ll  sum = 0;
    if(a[0]==a[1]&&a[1]==a[2]){
        ll  num = mp[a[0]];
        sum = (num*(num-1)*(num-2))/6;
    }
    if(a[1]!=a[2]){
        ll num = mp[a[2]];
       sum = num;
    }
    if(a[0]!=a[1]&&a[1]==a[2]){
        ll num = mp[a[2]];
       // num = 99999;
        sum = (num*(num-1))/2;
    }

    cout<<sum<<endl;
}