GGCURR

/* the gcd function used here is from geeksforgeeks.com*/
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<cstdio>

using namespace std;
#define MOD 100007

// code for extended euclid's algorithm
int gcd(int a, int b, int *x, int *y)
{

    if (a == 0)
    {
        *x = 0;
        *y = 1;
        return b;
    }

    int x1, y1;
    int gc = gcd(b%a, a, &x1, &y1);

    *x = y1 - (b/a) * x1;
    *y = x1;

    return gc;
}

int main()
{
  int t,n,a,b,num;
  float c;
  cin>>t;
  while(t--)
  {
    cin>>n>>a>>b;
    int x0,y0;//these are the coefficient of a and b respec
    int g1=gcd(a,b,&x0,&y0);

    int cash=0,digital=0;
    //cerr<<g1<<" "<<x0<<" "<<y0<<endl;
    for(int i=0;i<n;i++)
      {
        scanf("%d",&num);
        //if num is a multiple of a or b then it is possible
        if(num%a==0 || num%b==0)
          {
            cash++; continue;
          }
        if(num % g1==0)  //change can be made by cash
        {
          //redue a and b such that gcd(a,b)=1
          int ta=a/g1;
          int tb=b/g1;
          float tc=num/g1;
          //cout<<ta<<" "<<tb<<" "<<tc<<endl;
          //find whether positive solutions are there or not
          if(x0 >0 && y0 >0)cash++;
          else if(x0 <0 && y0 < 0 )
          digital++;
          else
          {
            //find the negetive number
            if(y0 < 0)
            {
              int mul=ceil((-1*y0*tc)/ta);
              if((tc*x0 - mul*tb)<=0)
                digital++;
              else
                cash++;


            }
            else
            {
              int mul=ceil((-1*x0*tc)/tb);
              if((tc*y0 - mul*ta)<=0)
                digital++;
              else
                cash++;

            }
          }
        }
        else
          digital++;



    }
      printf("%d %d\n",cash,digital);
  }
}