Codeforces Round #406 (Div. 2)C. Berzerk

C. Berzerk

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 s_1, 1, s_1, 2, ..., s_1, k1 — Rick's set.

The third line contains integer k₂ followed by k₂ distinct integers s_2, 1, s_2, 2, ..., s_2, k2 — Morty's set

1 ≤ k_i ≤ n - 1 and 1 ≤ s_i, 1, s_i, 2, ..., s_i, 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.

Examples

Input

5
2 3 2
3 1 2 3

Output

Lose Win Win Loop
Loop Win Win Win

Input

8
4 6 2 3 4
2 3 6

Output

Win Win Win Win Win Win Win
Lose Win Lose Lose Win Lose Lose
题目大意：Rick和Morty一起做游戏，他们分别有集合s1，s2,里面有k1,k2个数字。

现在在一个有1~n顺时针围城的圆环里，1号为黑洞色的山洞，2~n为植物深林，深林里有一个怪物。

Rick和Morty想把怪物移动到山洞里，他们可以使怪物顺时针移动x个单位，x为集合里的数字。

Rick和Morty轮流移动，最后把怪物移动到洞里算胜利。

问怪物分别在2~n，Rick和Morty分别处于先手，是否会有先手必胜、必败，或者处于循环。
题目解析：如果A无论怎么移动，使B下一步必胜，则A的现在状态为必败。
　　　　　如果A可以移动任意一步，使B下一步必败，则A现在的状态为必胜。

#include <bits/stdc++.h>

using namespace std;
const int N = 7005;
bool win[N][2], vis[N][2];
int k[2], n;
vector<int>a[2];
int cnt[N][2];

void slove(int Pre_point, int Player){
        vis[Pre_point][Player] = true;
        for(auto Mov_lenth: a[!Player]){
        int Now_point = (Pre_point-Mov_lenth+n)%n;

        if(Now_point == 0) continue;
        if(vis[Now_point][!Player]) continue;

        if(win[Pre_ponint][Player] && ++cnt[Now_point][!Player] == k[!Player])
            slove(Now_point, !Player);
        else if(!win[Pre_point][Player])
            win[Now_point][!Player] = true,
            slove(Now_point, !Player);
    }
}
int main()
{
    scanf("%d", &n);
    for(int i = 0; i < 2; i++){
        scanf("%d", &k[i]);
        for(int j = 0, v; j < k[i]; j++){
            scanf("%d", &v);
            a[i].push_back(v);
        }
    }

    slove(0, 0);
    slove(0, 1);
    for(int i = 0; i < 2; i++){
        for(int j = 1; j < n; j++){
            if(!vis[j][i]) printf("Loop ");
            else printf("%s", win[j][i] > 0? "Win ":"Lose ");
        }
        printf("
");
    }
    return 0;
}