void test_case(){
    ll x, y;
    cin >> x >> y;
    ll iter = 0;

    // 1. Algo : If you are somewhere on (x, y) then you may have come from (x, y - x) or (x - y, y).
    // 2. So in short have come from (x , y%x)  or (x%y, x).
    // 3. One last thing to take keep in mind is that if you reach somewhere (c, 0) or (0, c).
    // 4. Then the solution is only possible if c = 1. And since you have overcounted 1 operation. So ans = ans - 1.



    while(x && y){
        ll here;
        if(x > y){
            here = x/y;
            x %= y;
        }else{
            here = y/x;
            y %= x;
        }
        iter += here;
    }

    if(x == 1 || y == 1)
        cout << iter - 1<< endl;
    else
        cout << -1 << endl;
}