/*
题意：u,v,w队长，队员，队长留下两个队员可以回家，两个队员留下，队长回家
2-sat问题，把两个队员看成一个整体就变成一个简单2-sat问题了
*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
using namespace std;
#define N 6100
#define NN 5100
struct node {
int u,v,w,next;
}bian[NN*8];
int head[N],yong,low[N],dfn[N],belong[N],ans,top,index,stac[N],vis[N];
void init()
{
memset(head,-1,sizeof(head));
yong=index=ans=top=0;
memset(vis,0,sizeof(vis));
memset(dfn,0,sizeof(dfn));
}
void addedge(int u,int v)
{
bian[yong].v=v;
bian[yong].next=head[u];
head[u]=yong++;
}
void tarjan(int u)
{
low[u]=dfn[u]=++index;
stac[++top]=u;
vis[u]=1;
int i;
for(i=head[u]; i!=-1; i=bian[i].next)
{
int v=bian[i].v;
if(!dfn[v])
{
tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(vis[v])
low[u]=min(low[u],dfn[v]);
}
if(low[u]==dfn[u])
{
ans++;
int t;
do
{
t=stac[top--];
belong[t]=ans;
vis[t]=0;
}
while(t!=u);
}
}
int slove(int n)
{
int i;
for(i=0; i<=n; i++)
if(!dfn[i])
tarjan(i);
// printf("%d
",ans);
for(i=0; i<=n; i+=2)
if(belong[i]==belong[i+1])
return 0;
return 1;
}
int indx[N];
int main() {
int n,m,i,u,len,v,w;
while(scanf("%d%d",&n,&m)!=EOF) {
init();
len=-1;
for(i=0;i<n;i++) {
scanf("%d%d%d",&u,&v,&w);
indx[u]=++len;
indx[v]=++len;
indx[w]=len;
}
for(i=0;i<m;i++) {
scanf("%d%d",&u,&v);
addedge(indx[u],indx[v]^1);
addedge(indx[v],indx[u]^1);
}
if(!slove(len))
printf("no
");
else
printf("yes
");
}
return 0;}