#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1e6 + 7;
const int mod = 998244353;
#define ll __int128
//求解不定方程 ax+by=c 的最小整数解 x
ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if (!b) {x = 1; y = 0; return a;}
    ll ans = exgcd(b, a % b, x, y);
    ll tmp = x;
    x = y;
    y = tmp - a / b * y;
    return ans;
}
ll x, y;
ll cal(ll a, ll b, ll c)
{
    x = 0, y = 0;
    ll ans = exgcd(a, b, x, y);
    if (c % ans)return -1;
    x *= c / ans; b /= ans;
    // y *= c / ans;
    if (b < 0)b = -b;
    x = (x % b + b) % b;
    return x;
}
void solve()
{
    int a, b;
    cin >> a >> b;
    ll p = b, k = 1;
    while (p % 2 == 0) {
        p /= 2;
        k *= 2;
    }
    while (p % 5 == 0) {
        p /= 5;
        k *= 5;
    }
    long long c = 1e18, t = 0;
    auto check = [&](ll id) {
        c = cal(k, a, p * id);
        // cerr << c << "\n";
        t = (id*p - 1ll * k * c) / a;
        return t >= 1;
    };
    ll l = 1, r = 1e18;
    while (l < r) {
        ll mid = (l + r) >> 1;
        if (check(mid))r = mid;
        else l = mid + 1;
    }
    check(l);
    // cerr << (long long)l <<'\n';
    long long cx = c,tx = t;
    for(int i=1;i<=1000;i++){
        check(l + i);
        // cerr <<c <<'\n';
        if(c == 0)break;
        if(cx > c&&t*p<=1000000000ll&&t%2==0&&t%5==0){
            cx = c;
            tx = t;
        }
    }
    long long d = tx * p;
    // assert(d<=1000000000&&d>=1);
    cout << cx << " " << d << '\n';
}
int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int _ = 1;
    cin >> _;
    while (_--)
        solve();
    return 0;
}