How big can the answer be?

Which coordinates can you reach in 2 or 3 moves?

This problem is about handling cases.

The main solution is to quickly realize that $$$x \lt y$$$ can be done in 2 moves and $$$x \gt y+1$$$ with $$$y \gt 1$$$ in 3. You can guess that other cases are impossible.

Using two moves, we can move to any ($$$x$$$,$$$y$$$) with $$$x \lt y$$$ by first moving $$$x$$$ and then $$$y$$$. Using 3 moves $$$0 \lt a \lt b \lt c$$$ we get to ($$$a+c$$$,$$$b$$$) as follows: first to ($$$a$$$,$$$0$$$), then to ($$$a$$$, $$$b$$$), and then to ( $$$a+c$$$, $$$b$$$). Without loss of generality we can subtract $$$1$$$ from $$$a$$$ and add $$$1$$$ to $$$c$$$ anytime so we can assume $$$a=1$$$ and hence we can reach ($$$x$$$, $$$y$$$) in $$$3$$$ moves if and only if $$$x \gt y+1$$$ and $$$y \gt 1$$$.

Now assume we can reach some ($$$x$$$, $$$y$$$) with $$$k \gt 3$$$ jumps. The last two jumps can be merged with the previous 2 jumps giving a solution with $$$k-2$$$ jumps. Hence any other case is impossible.

#include <bits/stdc++.h>
#define endl '\n'
using namespace std;
#define int long long
 
signed main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int t;
	cin>>t;
	while(t--){
		int x,y;
		cin>>x>>y;
		if(x==y || x==y+1 || y==1)cout<<-1<<endl;
		else if(x<y)cout<<2<<endl;
		else cout<<3<<endl;
	}
}