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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define sz(x) ((int)x.size())
#define pb push_back
#define ii pair<int,int>
#define orta ((bas+son)/2)
#define st first
#define nd second
#define ll long long
#define N 200005
#define inf 1000000000
#define MOD 1000000007
using namespace std;

int n,m,q;
int dad[N];

int find(int x) {return dad[x]=(x==dad[x]?x:find(dad[x]));}

void fail() {cout<<"No";exit(0);}

int main() {

	ios_base::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);

	cin>>n>>m>>q;

	for(int i=1;i<=n;i++) dad[i]=i;

	int cnt=n;

	vector<ii> ch;

	for(int i=1;i<=q;i++) {

		int a,b,c;

		cin>>a>>b>>c;

		if(c==0) {

			if(find(a)!=find(b)) {

				dad[find(a)]=find(b);
				cnt--;

			}

		}
		else {

			ch.pb({a,b});

		}
 
	}

	for(auto x:ch) {

		if(find(x.st)==find(x.nd)) fail();

	}

	if(sz(ch) && cnt<3) fail();

	if((ll)cnt*(cnt+1)/2+n-cnt<m) fail();

	cout<<"Yes";

}