Rick and Morty are playing their own version of Berzerk (which has nothing in common with the famous Berzerk game). This game needs a huge space, so they play it with a computer.
In this game there are n objects numbered from 1 to n arranged in a circle (in clockwise order). Object number 1 is a black hole and the others are planets. There's a monster in one of the planet. Rick and Morty don't know on which one yet, only that he's not initially in the black hole, but Unity will inform them before the game starts. But for now, they want to be prepared for every possible scenario.
Each one of them has a set of numbers between 1 and n - 1 (inclusive). Rick's set is s1 with k1 elements and Morty's is s2 with k2 elements. One of them goes first and the player changes alternatively. In each player's turn, he should choose an arbitrary number like x from his set and the monster will move to his x-th next object from its current position (clockwise). If after his move the monster gets to the black hole he wins.
Your task is that for each of monster's initial positions and who plays first determine if the starter wins, loses, or the game will stuck in an infinite loop. In case when player can lose or make game infinity, it more profitable to choose infinity game.
Input
The first line of input contains a single integer n (2 ≤ n ≤ 7000) — number of objects in game.
The second line contains integer k1 followed by k1 distinct integers s1, 1, s1, 2, ..., s1, k1 — Rick's set.
The third line contains integer k2 followed by k2 distinct integers s2, 1, s2, 2, ..., s2, k2 — Morty's set
1 ≤ ki ≤ n - 1 and 1 ≤ si, 1, si, 2, ..., si, ki ≤ n - 1 for 1 ≤ i ≤ 2.
Output
In the first line print n - 1 words separated by spaces where i-th word is "Win" (without quotations) if in the scenario that Rick plays first and monster is initially in object number i + 1 he wins, "Lose" if he loses and "Loop" if the game will never end.
Similarly, in the second line print n - 1 words separated by spaces where i-th word is "Win" (without quotations) if in the scenario that Morty plays first and monster is initially in object number i + 1 he wins, "Lose" if he loses and "Loop" if the game will never end.
Example
52 3 23 1 2 3
Lose Win Win LoopLoop Win Win Win
84 6 2 3 42 3 6
Win Win Win Win Win Win WinLose Win Lose Lose Win Lose Lose 题目大意是：有n个位置1,2,3……n，围成1个圈，某个物体最开始的位置不在1，两个人轮流操作，每个人操作时可以让这个物体顺时针运动一些位置，使物体最终到达1号位置的人胜。求：物体初始在每个位置（不包括1），两个人分别先手的胜负情况。 感谢HX提供思路。。。 每个人每个状态无非就是三种情况：必胜（Win），必败（Lose），无法到达（Loop）。这其实是博弈论。 由于必败状态必定由所有必胜状态可推得，必胜状态只要1个必败状态就可以推出，那我们可以通过BFS/DFS的方式实现。设状态(x,y)表示当前是y操作，物体位置在x。那么(1,0)和(1,1)必然是必败状态。 假设我们使用BFS，当前状态为(ux,uy),下一个状态为(vx,vy)，那么事实上是由(vx,vy)推得(ux,uy)。但是我们知道的是最终状态，求的是初始状态，所以要反着来推。 如果(vx,vy)这个状态还没有确定，则： 如果(ux,uy)必败，(vx,vy)必胜； 如果(ux,uy)必胜，则要看看其他状态（同一层的）是否全部必胜，若是，则(vx,vy)必败。 代码如下：
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
;
struct node{
int x,f;
};
],a[][maxn],f[][maxn],cnt[][maxn];
int read(){
,f=; char ch=getchar();
'){if (ch=='-') f=-f; ch=getchar();}
+ch-',ch=getchar();
return x*f;
}
int main(){
n=read();
; i<; i++){
K[i]=read(); ; j<K[i]; j++) a[i][j]=read();
}
queue <node> Q; Q.push((node){,}); Q.push((node){,});
memset(f,,][]=f[][]=;
; i<; i++)
; j<=n; j++) cnt[i][j]=K[i];
for (; !Q.empty(); Q.pop()){
node u=Q.front(),v; v.f=-u.f;
; i<K[v.f]; i++){
v.x=u.x-a[v.f][i]; ) v.x+=n;
if (f[v.f][v.x]) continue;
) f[v.f][v.x]=,Q.push((node){v.x,v.f});
else{
cnt[v.f][v.x]--; ) f[v.f][v.x]=,Q.push((node){v.x,v.f});
}
}
}
; i<; i++,putchar('\n'))
; j<=n; j++) printf(??"Win":"Lose");
;
}