D. Power Products

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

You are given nn positive integers a1,…,ana1,…,an, and an integer k≥2k≥2. Count the number of pairs i,ji,j such that 1≤i<j≤n1≤i<j≤n, and there exists an integer xx such that ai⋅aj=xkai⋅aj=xk.

Input

The first line contains two integers nn and kk (2≤n≤1052≤n≤105, 2≤k≤1002≤k≤100).

The second line contains nn integers a1,…,ana1,…,an (1≤ai≤1051≤ai≤105).

Output

Print a single integer — the number of suitable pairs.

Example

input

Copy

6 3
1 3 9 8 24 1

output

Copy

5

Note

In the sample case, the suitable pairs are:

  • a1⋅a4=8=23a1⋅a4=8=23;
  • a1⋅a6=1=13a1⋅a6=1=13;
  • a2⋅a3=27=33a2⋅a3=27=33;
  • a3⋅a5=216=63a3⋅a5=216=63;
  • a4⋅a6=8=23a4⋅a6=8=23.

我写了很暴力的方法

分解质因子，然后计算出每个数要构成k次方数，所需的剩余因子个数

然后映射一个数量 每次查询map

用一个map<vector<pair<int,int> >,int>  来对应每个数所需要的因子和个数即可

#include<bits/stdc++.h>
using namespace std;
map<vector<pair<int,int> >, int>mp;
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    long long ans=0;
    for(int i=1;i<=n;i++)
    {
        vector<pair<int,int> >ve;
        ve.clear();
        int tmp,temp;
        scanf("%d",&tmp);
        ve.push_back(make_pair(1,1));
        temp=sqrt(tmp);
        for(int j=2;j<=temp;j++)
        {
            int num=0;
            while(tmp%j==0)
            {
                tmp/=j;
                num++;
            }
            num%=k;
            if(num!=0)
            {
                ve.push_back(make_pair(j,num));
                //cout<<j<<" "<<num<<endl;
            }
        }
        if(tmp!=1)
        {
            ve.push_back(make_pair(tmp,1));
            //cout<<tmp<<" "<<1<<endl;
        }
        ans+=mp[ve];
        for(int j=1;j<ve.size();j++)
        {
            ve[j].second=k-ve[j].second;
        }
        //cout<<ve[1].second<<endl;
        mp[ve]++;
        //cout<<ans<<endl;
    }
    printf("%I64d\n",ans);