#include <bits/stdc++.h>
using namespace std;
#define N 200005
typedef long long ll;
const ll e9 = 1000000000;
vector<ll> rec;
ll gcd(ll a, ll b) {
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

ll exgcd(ll a, ll b, ll& x, ll& y) {
    if (b == 0) {
        x = 1; y = 0;
        return a;
    }
    ll d = exgcd(b, a % b, x, y);
    ll temp = x; x = y; y = temp - y * (a / b);
    return d;
}

ll cal(ll a, ll b, ll c)
{
    ll x, y;
    ll gcd = exgcd(a, b, x, y);
    if (c % gcd != 0) return -1;
    x *= c / gcd;//转化为a*x+b*y=c的解
    b /= gcd;//约去c后原来b就变为了b/gcd;
    if (b < 0) b = -b;//如果b为负数就去绝对值
    ll ans = x % b;
    if (ans <= 0) ans += b;//求最小正整数解
    return ans;//返回的就是最小正整数解
}

void solve() {
    ll a, b;
    cin >> a >> b;
    ll w = b;
    while (w % 2 == 0)
        w /= 2;
    while (w % 5 == 0)
        w /= 5;
    ll minc = e9, ansd = e9;
    for (auto e : rec)
    {
        ll c, k;
        //aew = ww * k + b(-c)
        //ae = w * k + eb(-c)
        ll eb = b / w;
        k = cal(w, eb, a * e);
        ll de = (a * e - w * k) / w;
        if ((a * e - w * k) % w)
            de++;
        if (de > 0)
            k += de;
        ll d = w * e;
        c = (w * k - a * e) / eb;
        ll g_cd = gcd(c, d);
        c /= g_cd;
        d /= g_cd;
        if (c < minc && d <= e9) {
            minc = c;
            ansd = d;
        }
        else if (c == minc && d < ansd) {
            minc = c;
            ansd = d;
        }
    }
    cout << minc << " " << ansd << "\n";
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    int t = 1;
    cin >> t;
    ll x1 = 1;
    for (int i = 1; i <= 32; i++) {
        ll x2 = 1;
        for (int j = 1; j <= 20; j++){
            if (x1 > e9 || x2 > e9)
                continue;
            if (x1 * x2 > e9)
                continue;
            rec.push_back(x1 * x2);
            x2 *= 5;
        }
        x1 *= 2;
    }
    while(t--){
        solve();
    }
}