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>
const int N=100005;
int n,k,f[N];
long long m;
struct note{
	int x,y,c;
}a[N];
bool cmp(note x,note y){
	return x.c<y.c;
}
int find(int x){
	if (f[x]==x) return x;
	f[x]=find(f[x]);
	return f[x];
}
int main(){
	scanf("%d%lld%d",&n,&m,&k);
	for (int i=1;i<=k;i++){
		scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].c);
		if (a[i].c && m==n-1){
			puts("No");
			return 0;
		}
	} 
	for (int i=0;i<n;i++) f[i]=i;
	std::sort(a+1,a+k+1,cmp);
	for (int i=1;i<=k;i++){
		int fx=find(a[i].x),fy=find(a[i].y);
		if (a[i].c){	
			if (fx==fy){
				puts("No");
				return 0;
			}
		}else{
			f[fx]=fy;
		}
	}
	int cnt=0;
	if (m==n-1) {
		puts("Yes");return 0; 	
	}
	for (int i=0;i<n;i++) if (find(i)==i) cnt++;
	if (n-cnt+1ll*cnt*(cnt-1)/2<m || cnt==2 ) puts("No");
	else puts("Yes");
}