题解 | #小红的取模构造#

讨论。

#include<bits/stdc++.h>
#define int long long
#define double long double
#define x first
#define y second
using namespace std;
typedef long long LL;
typedef long long ll;
typedef pair<int, int> PII;
const int N = 3e5 + 10;
const int M = 1e3 + 10;
int mod = 1e9 + 7;
// int a[N];

void solve() {
    int a, b;
    cin >> a >> b;
    if (a == 0 && b == 0) {
        cout << 1 << " " << 1 << "\n";
        return ;
    }
    if (a == 0) {
        cout << 2 * b << " " << b << "\n";
        return ;
    }
    if (b == 0) {
        cout << a << " " << a * 2 << "\n";
        return ;
    }
    if (b > a) {
        cout << a + b << " " << b << "\n";
        return ;
    } else if (a > b) {
        cout << a << " " << a + b << "\n";
        return ;
    }
    cout << -1 << " " << -1 << "\n";
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int _;
    _ = 1;
    cin >> _;
    while (_--) {
        solve();
    }
}