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 rep(i,n)for(int i=0;i<(n);i++)
using namespace std;                                      
typedef long long ll;
typedef pair<int,int>P;

class UnionFind {
	vector<int>par, sz;
public:
	UnionFind() {}
	UnionFind(int n) {
		par = sz = vector<int>(n);
		for (int i = 0; i < n; i++) {
			par[i] = i;
			sz[i] = 1;
		}
	}
	int find(int x) {
		if (par[x] == x)return x;
		return par[x] = find(par[x]);
	}
	void unite(int x, int y) {
		x = find(x); y = find(y);
		if (x == y)return;
		if (sz[x] > sz[y]) {
			par[y] = x;
			sz[x] += sz[y];
		}
		else {
			par[x] = y;
			sz[y] += sz[x];
		}
	}
	bool same(int x, int y) {
		return find(x) == find(y);
	}
	int size(int x) {
		return sz[find(x)];
	}
};

int a[200000],b[200000],c[200000];

int main(){
	int n;ll m;int q;cin>>n>>m>>q;
	UnionFind uf(n);
	rep(i,q){
		scanf("%d%d%d",&a[i],&b[i],&c[i]);
		if(c[i]==0)uf.unite(a[i],b[i]);
	}
	rep(i,q){
		if(c[i]==1){
			if(uf.same(a[i],b[i])){
				puts("No");return 0;
			}
		}
	}
	int Min=0;
	int cnt=0;
	rep(i,n){
		if(uf.find(i)!=i)continue;
		cnt++;
		Min+=uf.size(i)-1;
	}
	if(cnt==1){
		if(m==Min)puts("Yes");
		else puts("No");
	}
	else if(cnt==2){
		puts("No");
	}
	else if(Min+cnt<=m&&m<=Min+cnt*ll(cnt-1)/2){
		puts("Yes");
	}
	else puts("No");
}