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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 100010
typedef long long ll;
#define NO !printf("No\n")
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,q,a[N],b[N],c[N];
ll m;
struct Union_set{
	int f[N];
	void init(int n){
		for(int i=0;i<n;++i){
			f[i]=i;
		}
	}
	inline int getf(int x){
		return f[x]==x?x:f[x]=getf(f[x]);
	}
	int Merge(int x,int y){
		int fx=getf(x),fy=getf(y);
		if(fx==fy)return 0;
		f[fy]=fx;
		return 1;
	}
}S;
int main(){
	n=read(),m=read(),q=read();
	for(int i=1;i<=q;++i){
		a[i]=read()+1,b[i]=read()+1,c[i]=read();
	}
	S.init(n);
	int tot=n;
	for(int i=1;i<=q;++i){
		if(!c[i]){
			tot-=S.Merge(a[i],b[i]);
		}
	}
	if(1LL*tot*(tot-1)/2+n-tot<m)return NO;
	
	int res=0;
	for(int i=1;i<=q;++i){
		if(c[i]){
			if(S.getf(a[i])==S.getf(b[i]))return NO;
		}
	}
	for(int i=1;i<=q;++i){
		if(c[i])res+=S.Merge(a[i],b[i]);
	}
	if(res){
		if(m<n)return NO;
	}
	printf("Yes\n");
	return 0;
}