#include<bits/stdc++.h>
using namespace std;
#define int long long
#define pll pair<int,int>
#define lf float
const int P = 998244353;
const int MX = 1e12;
int qpow(int a, int b) {
    int ans = 1;
    while (b) {
        if (b & 1)ans = ans * a % P;
        a = a * a % P;
        b >>= 1;
    }
    return ans;
}
int gcd(int a, int b) {
    int t;
    while (b) {
        t = b;
        b = a % b;
        a = t;
    }
    return a;
}
int T[100][100] = { 0 };
int K[10000], tk = 0;
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;
}
int ni(int a,int p){
    int x, y;
    exgcd (a, p, x, y);
    x = (x % p + p) % p;
    return x;
}
void solve() {
    int a, b;
    cin >> a >> b;
    int x = b, A = 1;
    while (x % 2 == 0)x /= 2, A *= 2;
    while (x % 5 == 0)x /= 5, A *= 5;
    if (x == 1) {
        cout << 0 << ' ' << 1 << '\n';
        return;
    }
    a %= x;
    int minc = 1e12, mind = 1e12;
    for (int i = 1; i <= tk; i++) {
        int c = (x - K[i] * a % x) % x;
        int d = K[i] * x * A;
        int g = gcd(c, d);
        c /= g, d /= g;
        if (d <= 1000000000 && c < minc) {
            minc = c;
            mind = d;
        }
    }
    for (int i = 1; i <= tk; i++) {
        if(A%K[i])continue;
        int tx,ty;
        exgcd(K[i],x,tx,ty);
        tx*=a;
        ty*=a;
        int c=(tx%x+x)%x;
        int d =  x * A/K[i];
        int g = gcd(c, d);
        c /= g, d /= g;
        if (d <= 1000000000 && c < minc) {
            minc = c;
            mind = d;
        }
    }
    cout << minc << " " << mind << '\n';
}
signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    T[0][0] = 1;
    for (int i = 0; i < 55; i++) {
        for (int j = 0; j < 55; j++) {
            if (T[i][j] > MX)break;
            T[i + 1][j] = T[i][j] * 2;
            T[i][j + 1] = T[i][j] * 5;
        }
    }
    for (int i = 0; i < 55; i++) {
        for (int j = 0; j < 55; j++) {
            if (T[i][j] > MX)break;
            if (T[i][j] == 0)break;
            K[++tk] = T[i][j];
        }
    }
    int t = 1;
    cin >> t;
    while (t--)
        solve();
}