#include<bits/stdc++.h>
using namespace std;
#define int long long
#define gcd __gcd
const int maxd=1E9;

template <class T> T sign(const T &a) {
    return a == 0 ? 0 : (a < 0 ? -1 : 1);
}
template <class T> T ceil(const T &a, const T &b) {
    T A = abs(a), B = abs(b);
    assert(b != 0);
    return sign(a) * sign(b) > 0 ? (A + B - 1) / B : -A / B;
}
int exgcd(int a, int b, int &x, int &y) {
    if (!b) {
        x = 1, y = 0;
        return a;
    }
    int d = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}

void solve()
{
    int a,b,c=1E18,d=0,c1=0,c2=0;
    scanf("%lld %lld",&a,&b);
    int k=b;
    while(k%2==0){k=k/2;c1++;}
    while(k%5==0){k=k/5;c2++;}
    if(k==1)
    {
        printf("0 1\n");return;
    }
    map<int,int> mp;
    priority_queue<int,vector<int>,greater<int>> pq;
    mp[b]=1;
    pq.push(b);
    while(!pq.empty())
    {
        int x=pq.top();pq.pop();
        if(5*x<=maxd&&!mp[5*x])
        {
            mp[5*x]=1;
            pq.push(5*x);
        }
        if(2*x<=maxd&&!mp[2*x])
        {
            mp[2*x]=1;
            pq.push(2*x);
        }
        int l=0,r=1E18/k;
        int y=a*x/b;
        while(l<r)
        {
            int mid=(l+r)/2;
            if(mid*k<y){l=mid+1;}
            else{r=mid;}
        }
        while(l*k<y){l++;}
        int o=l*k-y;
        int m=gcd(o,x);
        o=o/m;x=x/m;
        if(o<c){c=o;d=x;}
    }


    auto clac = [&](int a, int b, int c) -> int {
        int x, y, d = exgcd(a, b, x, y);
        if (c % d != 0) {
            return -1;
        }
        x *= c / d, y *= c / d;
        int p = b / d, q = a / d, k;
        if (x < 0) {
            k = ceil(1 - x, p);
            x += p * k;
            y -= q * k;
        }
        else if (x >= 0) { //将x提高到最小正整数
            k = (x - 1) / p;
            x -= p * k; //将x降低到最小正整数
            y += q * k;
        }
        if (y > 0) { //有正整数解
            return x; 
        } else { //无整数解
            return -1;
        }
    };

    for(int i=0;i<=c1;i++)
    {
        for(int j=0;j<=c2;j++)
        {
            int x=k*pow(2,i)*pow(5,j);
            int y=b/x;
            // a/b+c/x  y=b/x
            // a = k*m-c*x
            int res=clac(k,-y,a);
            if(res!=-1&&res<c)
            {
                c=res;
                d=x;
            }
        }
    }

    printf("%lld %lld\n",c,d);
}

signed main()
{
    int tt;
    scanf("%lld",&tt);
    while(tt--)
    {
        solve();
    }
}