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 LL long long
using namespace std;
const int N=200010;
int n,m,k,fa[N],is[N],tot;
struct Q{
	int u,v,id;
	bool operator<(const Q&o)const{
		return id<o.id;
	}
}q[N];
int getfa(int x){
	return fa[x]==x?x:fa[x]=getfa(fa[x]);
}
int main(){
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=k;i++){
		scanf("%d%d%d",&q[i].u,&q[i].v,&q[i].id);
		q[i].u++,q[i].v++;
	}
	sort(q+1,q+k+1);
	for(int i=1;i<=n;i++){
		fa[i]=i;
	}
	for(int i=1;i<=k;i++){
		if(q[i].id==0){
			int u=getfa(q[i].u),v=getfa(q[i].v);
			if(u!=v)fa[u]=v,m--;
		}
		else{
			int u=getfa(q[i].u),v=getfa(q[i].v);
			if(u==v){
				puts("No");
				return 0;
			}
			is[u]=is[v]=1;
		}
	}
	bool flag=0;
	for(int i=1;i<=n;i++){
		if(fa[i]==i){
			if(is[i]==0)flag=1;
			tot++;
		}
	}
	LL mi,mx;
	if(flag)mi=tot-1;
	else mi=tot;
	mx=(LL)tot*(tot-1)/2;
	if(mi<=m&&m<=mx)puts("Yes");
	else puts("No");
}