Problem: 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`.

Examples

Input

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


Output

Yes


Input

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


Output

No


Input

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


Output

No

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define int long long  
using namespace std;

inline int read()
{
	int sum=0,ff=1; char ch=getchar();
	while(!isdigit(ch))
	{
		if(ch=='-') ff=-1;
		ch=getchar();
	}
	while(isdigit(ch))
		sum=sum*10+(ch^48),ch=getchar();
	return sum*ff;
}


const int N=1e5+5;

int n,m,fa[N],tot,al,T,ask;

struct nood
{
	int u,v,op;
};
nood e[N<<1];

inline int find(int x)
{
	if(x==fa[x]) return x;
	return fa[x]=find(fa[x]);
}

inline void sub_tree()
{
	if(m<n-1)
		printf("No\n");
	if(ask==T) 
		printf("Yes\n");
	else printf("No\n");
	exit(0);
}

signed main()
{
	n=read();
	m=read();
	T=read();
	for ( int i=0;i<=n;i++ ) fa[i]=i;
	for ( int i=1;i<=T;i++ )
	{
		e[i]=(nood){read(),read(),read()};
		if(!e[i].op)
		{
			ask++;
			int x=find(e[i].u),y=find(e[i].v);
			if(x==y) continue;
			al++; 
			fa[x]=y;
		}
	}
	if(m<=n-1) sub_tree();
	for ( int i=1;i<=T;i++ ) 
	{
		if(e[i].op)
		{
			int x=find(e[i].u),y=find(e[i].v);
			if(x==y) return printf("No\n"),0;
		}
	}
	tot=n-al;
	int l=n,r=al+(tot*(tot-1))/2;
//	cout<<m<<" "<<r<<endl;
	if(m>=l&&m<=r) printf("Yes\n");
	else printf("No\n");
	return 0;
}