InputInput contains multiple sets of data.For each set of data,there are four numbers in the first line:N (1 ≤ N≤ 1000)、M(1 ≤M ≤ 10000)、X,Y(X+Y≤N ),in order to show the number of players(numbered 1toN ),the number of matches,the number of known "good players" and the number of known "bad players".In the next M lines,Each line has two numbersa, b(a≠b) ,said there is a game between a and b .The next line has X different numbers.Each number is known as a "good player" number.The last line contains Y different numbers.Each number represents a known "bad player" number.Data guarantees there will not be a player number is a good player and also a bad player.OutputIf all the people can be divided into "good players" and "bad players”, output "YES", otherwise output "NO".Sample Input
5 4 0 0
1 3
1 4
3 5
4 5
5 4 1 0
1 3
1 4
3 5
4 5
2
Sample Output
NO
YES
Hint
/*
* @Author: lyuc
* @Date: 2017-05-01 15:48:50
* @Last Modified by: lyuc
* @Last Modified time: 2017-05-01 20:33:47
*/
/**
* 题意：有n个人每个人只能是好人或者是坏人，给你n对人的关系，每对的中两个人的关系是对立的，一定有一个
* 是坏人一个是好人，并且给了 x个确定是好人的编号， y个确定是坏人的编号，现在让你判断，是否能将
* 所有人划分成两个阵营（有人的身份不能确定也不行）
*
* 思路：裸的二分染色，按照输入情况进行建边，然后按照输入的x,y进行染色判断如果有矛盾那一定是不行的，
* 最后在讲给出的条件进行染色，最后如果有身份不明的人就可以除去了
*/
#include <stdio.h>
#include <vector>
#include <string.h>
#include <queue>
#include <iostream>
using namespace std;
int n,m;
int x,y;
int a[],b[];
vector<int>edge[];
int vis[];
bool ok;
int str;
void bfs(int u,int flag){
queue<int>q;
q.push(u);
vis[u]=flag;
while(!q.empty()){
int tmp=q.front();
q.pop();
for(int i=;i<edge[tmp].size();i++){
int v=edge[tmp][i];
if(vis[v]==vis[tmp]){
ok=false;
return;
}
if(vis[v]==){
vis[v]=(-vis[tmp]);
q.push(v);
}
}
}
}
void init(){
ok=true;
memset(vis,,sizeof vis);
for(int i=;i<;i++){
edge[i].clear();
}
}
int main(){
// freopen("in.txt","r",stdin);
while(scanf("%d%d%d%d",&n,&m,&x,&y)!=EOF){
init();
for(int i=;i<m;i++){
scanf("%d%d",&a[i],&b[i]);
edge[a[i]].push_back(b[i]);
edge[b[i]].push_back(a[i]);
}
for(int i=;i<x;i++){
scanf("%d",&str);
if(vis[str]==-){
ok=false;
}
bfs(str,);
}
for(int i=;i<y;i++){
scanf("%d",&str);
if(vis[str]==){
ok=false;
}
bfs(str,-);
}
for(int i=;i<m;i++){
if(vis[a[i]]==&&vis[b[i]]==)
bfs(a[i],);
}
for(int i=;i<=n;i++){
if(vis[i]==){
ok=false;
break;
}
}
printf(ok?"YES\n":"NO\n");
}
return ;
}