#include<bits/stdc++.h>
using namespace std;
const int N=2100;
vector<int>G[N];
int tsp, cnt, dfn[N], low[N], scc[N];
stack<int>stk;bool instk[N];
void tarjan(int x)
{
    dfn[x]=low[x]=++tsp;
    stk.push(x);instk[x]=1;
    for(int y:G[x])
    {
        if(dfn[y]==0)
        {
            tarjan(y);
            low[x]=min(low[x], low[y]);
        }
        else if(instk[y])low[x]=min(low[x], dfn[y]);
    }
    if(low[x]==dfn[x])
    {
        cnt++;
        for(int z=-1;z!=x;)
        {
            z=stk.top();stk.pop();instk[z]=0;
            scc[z]=cnt;
        }
    }
}
int main()
{
    int n, m;
    while(scanf("%d", &n) && n)
    {
        scanf("%d", &m);
        memset(G, 0, sizeof(G));
        for(int i=1;i<=m;i++)//构图（将点的关系简单梳理了一下）
        {
            int a1, a2, c1, c2;scanf("%d%d%d%d", &a1, &a2, &c1, &c2);//a1+c1*n 和 a2+c2*n有矛盾 
            G[a1+c1*n].push_back(a2+(c2^1)*n);
            G[a2+c2*n].push_back(a1+(c1^1)*n);
        }
        tsp=cnt=0;memset(dfn, 0, sizeof(dfn));memset(low, 0, sizeof(low));
		memset(instk, 0, sizeof(instk));memset(scc, 0, sizeof(scc));
        for(int i=1;i<=2*n;i++)if(dfn[i]==0)tarjan(i);

        bool bk=1;
        for(int i=1;i<=n;i++)if(scc[i]==scc[i+n]){bk=0;break;}//夫妻不能同时选
        if(!bk)printf("NO\n");else printf("YES\n");
    }
    return 0;
}