Problem:
Score : 700 points

Problem Statement

Snuke's mother gave Snuke an undirected graph consisting of N vertices numbered 0 to N-1 and M edges.
This graph was connected and contained no parallel edges or self-loops.

One day, Snuke broke this graph.
Fortunately, he remembered Q clues about the graph.
The i-th clue (0 \leq i \leq Q-1) is represented as integers A_i,B_i,C_i and means the following:

If C_i=0: there was exactly one simple path (a path that never visits the same vertex twice) from Vertex A_i to B_i.

If C_i=1: there were two or more simple paths from Vertex A_i to B_i.

Snuke is not sure if his memory is correct, and worried whether there is a graph that matches these Q clues.
Determine if there exists a graph that matches Snuke's memory.

Constraints

2 \leq N \leq 10^5

N-1 \leq M \leq N \times (N-1)/2

1 \leq Q \leq 10^5

0 \leq A_i,B_i \leq N-1

A_i \neq B_i

0 \leq C_i \leq 1

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M Q
A_0 B_0 C_0
A_1 B_1 C_1
\vdots
A_{Q-1} B_{Q-1} C_{Q-1}

Output

If there exists a graph that matches Snuke's memory, print Yes; otherwise, print No.

Sample Input 1

5 5 3
0 1 0
1 2 1
2 3 0

Sample Output 1

Yes

For example, consider a graph with edges (0,1),(1,2),(1,4),(2,3),(2,4). This graph matches the clues.

Sample Input 2

4 4 3
0 1 0
1 2 1
2 3 0

Sample Output 2

No

Sample Input 3

10 9 9
7 6 0
4 5 1
9 7 0
2 9 0
2 3 0
4 1 0
8 0 0
9 1 0
3 0 0

Sample Output 3

No

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <cstdio>
#include <cstring>
#include <vector>
#define rep(a,b,c) for(int a=b;a<=c;a++)
#define pb push_back
using std::vector;
const int maxn=1e5+1;
#define int long long
inline int max(int a,int b) { return a>b?a:b; }
int vis[maxn],num,bel[maxn],len[maxn],n,m,q,tag[maxn],a[maxn],b[maxn],c[maxn],edge,mine,maxe,nans,sum,spj;
vector<int> e[maxn];
inline void dfs(int x) {
	len[num]++;  vis[x]=1; bel[x]=num;
	for(int v:e[x]) {
		if(vis[v]) continue;
		dfs(v);
	}
}
inline void check(int x,int rt) {
	if(nans) return;
	vis[x]=1;
	for(int v:e[x]) {
		if(bel[v]==bel[rt]) return(void)(nans=1);
		if(bel[v]) continue;
		if(!vis[v]) dfs(v);
	}
}
signed main() {
	scanf("%lld %lld %lld",&n,&m,&q);
	rep(i,1,q) {
		scanf("%lld %lld %lld",&a[i],&b[i],&c[i]);
		a[i]++; b[i]++;
		if(c[i]) e[a[i]].pb(b[i]),e[b[i]].pb(a[i]),tag[a[i]]=1,tag[b[i]]=1,spj=1;
		else edge++;
	}
	if(!spj&&(m>=n)) { puts("No"); return 0; }
	rep(i,1,n) {
		if(vis[i]||e[i].empty()) continue;
		num++; dfs(i); len[num]=max(len[num],3ll);
		mine+=len[num],maxe+=((len[num]*(len[num]-1))/2); 
	}
	rep(i,1,n) e[i].clear(); memset(vis,0,sizeof(vis));
	rep(i,1,num) maxe+=(sum*len[num]),sum+=len[num];
	rep(i,1,q) {
		if(!c[i]) e[a[i]].pb(b[i]),e[b[i]].pb(a[i]);
	}
	rep(i,1,n) {
		if(tag[i]) check(i,i);
		if(nans) break;
	}
	mine+=edge; maxe+=edge;
	if(nans) puts("No");
	else if(mine<=m&&maxe>=m) puts("Yes");
	else puts("No");
}