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>
using namespace std;
template<class T> T read(){
	T x=0,w=1;char c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-') w=-w;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*w;
}
template<class T> T read(T&x){
	return x=read<T>();
}
#define CO const
#define IN inline
typedef long long LL;

CO int N=100000+10;
struct edge {int u,v,c;}e[N];
IN bool operator<(CO edge&a,CO edge&b){
	return a.c<b.c;
}

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

bool vis[N];

int main(){
	int n=read<int>();
	LL m=read<LL>();
	int q=read<int>();
	for(int i=1;i<=q;++i){
		int u=read<int>()+1,v=read<int>()+1;
		e[i]=(edge){u,v,read<int>()};
	}
	sort(e+1,e+q+1);
	for(int i=1;i<=n;++i) fa[i]=i;
	bool valid=1,more=1;
	for(int i=1;i<=q;++i){
		if(e[i].c==0){
			int fu=find(e[i].u),fv=find(e[i].v);
			if(fu!=fv) fa[fu]=fv;
		}
		else{
			more=0;
			int fu=find(e[i].u),fv=find(e[i].v);
			if(fu==fv) {valid=0;break;}
		}
	}
	if(!valid) return puts("No"),0;
	int cnt=0;
	for(int i=1;i<=n;++i) cnt+=find(i)==i;
	puts(n-more<=m and m<=n+(LL)cnt*(cnt-1)/2-cnt?"Yes":"No");
	return 0;
}