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<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int N=100005;

int n,m,q,ecnt,cnt1;
int fa[N];
vector<int>A,B;
int find(int x) {
	if(x==fa[x]) return x;
	return fa[x]=find(fa[x]);
}
int main() {
	scanf("%d %d %d",&n,&m,&q);
	for(int i=1;i<=n;i++) fa[i]=i;
	for(int i=1,u,v,c;i<=q;i++) {
		scanf("%d %d %d",&u,&v,&c);
		if(c==1) {
			A.push_back(u);
			B.push_back(v);
		}
		else {
			int fu=find(u);
			int fv=find(v);
			if(fu!=fv)
				fa[fu]=fv,ecnt++;
		}
	}
	for(int i=0;i<(int)A.size();i++) {
		int fu=find(A[i]);
		int fv=find(B[i]);
		if(fu==fv) {
			printf("No\n");
			return 0;
		}
	}
	long long l=n-ecnt;
	if(A.size()==0) {
		if(m<=ecnt+l*(l-1)/2)
			printf("Yes\n");
		else printf("No\n");
	}
	else {
		if(n<=m&&m<=ecnt+l*(l-1)/2) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}