//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>
2
3 using namespace std;
4
5 const int maxn = 100001;
6
7 int t;
8
9 int n;
10
11 int icase;
12
13 int a[maxn+10];
14
15 int b[maxn+10];
16
17 int main(){
18 scanf("%d",&t);
19 while(t--){
20 scanf("%d",&n);
21 memset(a, 0, sizeof(a));
22 memset(b, 0, sizeof(b));
23 for(int i = 1; i <= n; ++i){
24 scanf("%d",&a[i]);
25 }
26 for(int i = 1; i <= n; ++i){
27 scanf("%d",&b[i]);
28 }
29 if(n == 1){
30 if(a[1] >= b[1]){
31 printf("Case %d: Possible from station %d\n",++icase,1);
32 }else {
33 printf("Case %d: Not possible\n",++icase);
34 }
35 continue;
36 }
37 int cur = 1;
38 int i = 1;
39 int cnt = 0;
40 int fuel = 0;
41 int flag = 0;
42 while(cnt != n){
43 fuel += a[i];
44 if(fuel >= b[i]){
45 fuel -= b[i];
46 ++i;
47 ++cnt;
48 if(i > n){
49 i = 1;
50 flag = 1;
51 }
52 } else {
53 if(flag == 1){
54 flag = 2;
55 break;
56 }
57 ++i;
58 if(i > n){
59 i = 1;
60 flag = 1;
61 }
62 cur = i;
63 cnt = 0;
64 fuel = 0;
65 }
66 }
67 if(flag == 2){
68 printf("Case %d: Not possible\n",++icase);
69 } else {
70 printf("Case %d: Possible from station %d\n",++icase,cur);
71 }
72 }
73 }
