Flipping Frustration

#include <bits/stdc++.h>
using namespace std;
#define PB push_back
#define ZERO (1e-10)
#define INF (1<<29)
#define CL(A,I) (memset(A,I,sizeof(A)))
#define DEB printf("DEB!\n");
#define D(X) cout<<"  "<<#X": "<<X<<endl;
#define EQ(A,B) (A+ZERO>B&&A-ZERO<B)
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef pair<int,int> ii;
typedef vector<ii> vii;
#define IN(n) int n;scanf("%d",&n);
#define FOR(i, m, n) for (int i(m); i < n; i++)
#define REP(i, n) FOR(i, 0, n)
#define F(n) REP(i, n)
#define FF(n) REP(j, n)
#define FT(m, n) FOR(k, m, n)
#define aa first
#define bb second
void ga(int N,int *A){F(N)scanf("%d",A+i);}
#define MX (10000002)
bitset<MX> C;
int q[MX],qp[MX],b,e,N,L,R,T,p,u;
void bfs(){
    C.reset(),C[*q=1]=1,b=-1,e=1;
    if(T==1)return (void)puts("0");
    while(++b<e){
        u=q[b],p=qp[b];
        if(u-L>1&&!C[u-L]){
            if(u-L==T)return (void)printf("%d\n",p+1);
            C[q[e]=u-L]=1,qp[e++]=p+1;
        }
        if(u+R<=N&&!C[u+R]){
            if(u+R==T)return (void)printf("%d\n",p+1);
            C[q[e]=u+R]=1,qp[e++]=p+1;
        }
    }
    puts("uh-oh!");
}
int main(void){
    IN(_)F(_)scanf("%d%d%d%d",&N,&L,&R,&T),bfs();
    return 0;
}