/*
把起点和终点比作加油站，那总共有n+2个加油站了，
每次都求出从第0个到第j个加油站（j<i）分别在加满油的情况下到第i个加油站的最短时间dp[i]，
最终的dp[n+1]就是最优解了。
*/
# include <stdio.h>
# include <algorithm>
# include <string.h>
# define INF 999999999;
using namespace std;
int main()
{
    int L,n,c,t,i,j;
    double min1,tmp;
    int vr,vt1,vt2;
    int path[100010];
    double dp[100010];
    while(~scanf("%d%d%d%d",&L,&n,&c,&t))
    {
        scanf("%d%d%d",&vr,&vt1,&vt2);
        path[0]=0;
        for(i=1; i<=n; i++)
            scanf("%d",&path[i]);
        path[n+1]=L;
        dp[0]=0;
        for(i=1; i<=n+1; i++)
        {
            dp[i]=INF;
            for(j=0; j<i; j++)
            {
                int ll=path[i]-path[j];
                if(c>=ll)
                    tmp=ll*1.0/vt1;
                else
                    tmp=c*1.0/vt1+(ll-c)*1.0/vt2;
                if(j!=0)
                    tmp+=t;
                if(dp[i]>dp[j]+tmp)
                    dp[i]=dp[j]+tmp;
            }
        }
        if(dp[n+1]>L*1.0/vr)
            printf("Good job,rabbit!\n");
        else
            printf("What a pity rabbit!\n");

    }
    return 0;
}

hdu 2059 龟兔赛跑 (dp)