#include<bits/stdc++.h>
using namespace std;
#define int long long
int T;
int exgcd(int a,int b,int &x,int &y){
if(b==0){
x=1,y=0;
return a;
}
int d=exgcd(b,a%b,y,x);
y-=a/b*x;
return d;
}
signed main(){
cin>>T;
while(T--){
int n,d,x,y;
cin>>n>>d>>x>>y;
int a,b;
int gcd=exgcd(d,n,a,b);
if((y-x)%gcd)cout<<"Impossible\n";
else {
//把通解a0扩大为（y-x)/gcd
a=a*(y-x)/gcd;
//特解为a=a0+kn a最小且非负 所以为a=a0%kn
n/=gcd;//b扩大 但是n不扩大 n会缩小
cout<<(a%n+n)%n<<"\n";//保证非负
}
}
return 0;
}
