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;
int fa[100001];
int find(int x){return (fa[x]==x)?fa[x]:fa[x]=find(fa[x]);}
int a[100001],b[100001],c[100001];
signed main()
{
	int n,m,q;
	scanf("%d%d%d",&n,&m,&q);
	int num=n;
	for(int i=1;i<=n;i++)
		fa[i]=i;
	for(int i=1;i<=q;i++)
	{
		scanf("%d%d%d",&a[i],&b[i],&c[i]);
		if(c[i]==0)
		{
			int f1=find(a[i]),f2=find(b[i]);
			if(f1!=f2)
			{
				fa[f1]=f2;
				num--;
			}
		}
	}
	for(int i=1;i<=q;i++)
	{
		if(c[i]==1)
		{
			if(m<n||num<=2||find(a[i])==find(b[i]))
			{
				printf("No");
				return 0;
			}
		}
	}
	if(m-(n-num)>1ll*(num-1)*num/2)
	{
		printf("No");
		return 0;
	}
	printf("Yes");
	return 0;
}