#include<bits/stdc++.h>
#define int long long
using namespace std;

const int N = 1e5+10;

int num[N],p[N],cnt;

long long quik_power(int base, int power)
{
	long long result = 1;   
	while (power > 0)           
	{
		if (power & 1)         							
			result = result * base;   
			//根据公式每个项都取余数后在再做累乘
		base = base * base ;   
			//根据公式每个项都取余数后在再做平方操作      						
		power >>= 1;         						
	}
			//根据公式在最后的的结果上再来一次取余数
	return result;       
}

void solve()
{
    int c = 0x3f3f3f3f3f3f3f3f;
    int a,b;cin >> a >> b;
    int x = b;
    int ansd=0;
    while(x%2==0)x/=2;
    while(x%5==0)x/=5;
    //a/b+c/d
    //ad+bc = kbd
    //c = (kbd-ad)/b
    //k*bd-ad>=0;
    //(k*bd-ad)/b = k*d-a(2^i*5^j)>=0
    //(k*b-a)(d/b);
    //k>=a/b
    //1<=1/k <=b/a

    for(int i = 1;i<=cnt;i++)
    {
        int d = num[i]*x;
        if(d>1e9)break;
        int j = lower_bound(num+1,num+1+cnt,b/a)-num;
        int k = num[j];
        // cout << d << ' ' << k << '|' << endl;
        if(a*d%b||d%k)continue;
        if(c>d/k-a*d/b)c = d/k-a*d/b,ansd = d;
    }
    if(c==0){cout << 0 << ' ' << 1 << endl;return;}
    cout << c << ' ' << ansd << endl;

}

signed main()
{
    int tt =1;
    cin >> tt;
    for(int i = 0;i<=29;i++)
    {
        for(int j = 0;j<=29;j++)
        {
            int cc = quik_power(2,i)*quik_power(5,j);
            if(cc<=1e9)
            {
                num[++cnt] = cc;
            }
            else break;
        }
    }
    sort(num+1,num+1+cnt);
    // cout << cnt << endl;
    while(tt--)
    {
        solve();
    }
}