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;
#define int long long
#define y1 ___y1
const int N=200005;
map<int ,int > Map[N];
int n,m,Q,opt,x,y,f1[N],X[N],Y[N],tot,f2[N],size[N];
int find1(int x){
	if (x==f1[x])return x;
	return f1[x]=find1(f1[x]);
}
int find2(int x){
	if (x==f2[x])return x;
	return f2[x]=find2(f2[x]);
}
signed main(){
	scanf("%lld%lld%lld",&n,&m,&Q);
	int flag=0,cnt=0;
	for (int i=1;i<=n;i++)f1[i]=f2[i]=i,size[i]=1;
	while (Q--){
		scanf("%lld%lld%lld",&x,&y,&opt);
		x++;y++;
		if (opt==0){
			int x1=find1(x),y1=find1(y);
			if (x1!=y1){
				f1[x1]=y1;
				size[y1]+=size[x1];
				cnt++;
			}
		}
		else {
			X[++tot]=x;
			Y[tot]=y;
			flag=1;
		}
	}
	for (int i=1;i<=tot;i++)
		if (find1(X[i])==find1(Y[i])){
			puts("No");
			return 0;
		}
	if (m==n-1&&flag){
		puts("No");
		return 0;
	}
	if (m>(n-cnt)*(n-cnt-1)/2+cnt){
		puts("No");
		return 0;
	}
	puts("Yes");
}