//当汽车从第i个加油站到第j个加油站无法继续走下去的时候，这时候[i,j]区间的所有加油站都无法作为起点，因为当我们到第k个加油站的时候，起码是带着>=0的油去的，现在不带油直接从第k个开始肯定更不行了。
 1 #include<bits/stdc++.h>

 using namespace std;

 const int maxn = ;

 int t;

 int n;

 int icase;

 int a[maxn+];

 int b[maxn+];

 int main(){

         scanf("%d",&t);

         while(t--){

             scanf("%d",&n);

             memset(a, , sizeof(a));

             memset(b, , sizeof(b));

             for(int i = ; i <= n; ++i){

                 scanf("%d",&a[i]);

             }

             for(int i = ; i <= n; ++i){

                 scanf("%d",&b[i]);

             }

             if(n == ){

                 if(a[] >= b[]){

                     printf("Case %d: Possible from station %d\n",++icase,);

                 }else {

                     printf("Case %d: Not possible\n",++icase);

                 }

                 continue;

             }

             int cur = ;

             int i = ;

             int cnt = ;

             int fuel = ;

             int flag = ;

             while(cnt != n){

                 fuel += a[i];

                 if(fuel >= b[i]){

                     fuel -= b[i];

                     ++i;

                     ++cnt;

                     if(i > n){

                         i = ;

                         flag = ;

                     }

                 } else {

                     if(flag == ){

                         flag = ;

                         break;

                     }

                     ++i;

                     if(i > n){

                         i = ;

                         flag = ;

                     }

                     cur = i;

                     cnt = ;

                     fuel = ;

                 }

             }

             if(flag == ){

                 printf("Case %d: Not possible\n",++icase);

             } else {

                 printf("Case %d: Possible from station %d\n",++icase,cur);

             }

         }

 }