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: Accepted)

#include<cstdio>
#include<cstdlib>
#include<cassert>
#include<iostream>
#define llong long long
using namespace std;

inline int read()
{
	int x=0; bool f=1; char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=0;
	for(; isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^'0');
	if(f) return x;
	return -x;
}

const int N = 2e5;
struct Element
{
	int u,v,typ;
} a[N+3];
int uf[N+3];
int n,q,c; llong m;

int findfa(int u) {return uf[u]==u?u:uf[u]=findfa(uf[u]);}

int main()
{
	scanf("%d%lld%d",&n,&m,&q);
	for(int i=1; i<=n; i++) uf[i] = i;
	for(int i=1; i<=q; i++)
	{
		scanf("%d%d%d",&a[i].u,&a[i].v,&a[i].typ); a[i].u++; a[i].v++;
		if(a[i].typ==0)
		{
			int uu = findfa(a[i].u),vv = findfa(a[i].v);
			if(uu!=vv) {uf[uu] = vv;}
		}
	}
	int cnt1 = 0; c = 0;
	for(int i=1; i<=n; i++) {if(i==findfa(i)) c++;}
	for(int i=1; i<=q; i++)
	{
		if(a[i].typ==1)
		{
			cnt1++;
			int uu = findfa(a[i].u),vv = findfa(a[i].v);
			if(uu==vv) {puts("No"); return 0;}
		}
	}
	llong l = cnt1?max(3ll,(llong)c):c-1ll,r = 1ll*c*(c-1ll)/2ll;
	if(m-(n-c)>=l && m-(n-c)<=r) {puts("Yes");}
	else {puts("No");}
	return 0;
}