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>
#define int long long
using namespace std;
const int maxn=1000005;
int a[maxn],b[maxn],c[maxn],n,m,q,f[maxn],mi,ma,tot[maxn];
int find(int x)
{
 return x==f[x]?x:f[x]=find(f[x]);
}
int max(int x,int y)
{
	return x>y?x:y;
}
signed main()
{
	scanf("%lld%lld%lld",&n,&m,&q);
	for(int i=1;i<=n;++i)f[i]=i;int flag=0;
	for(int i=1;i<=q;++i)
	{
		 scanf("%lld%lld%lld",&a[i],&b[i],&c[i]);a[i]++;b[i]++;
		 if(!c[i])
		 {
		 	int x=find(a[i]);int y=find(b[i]);
		 	if(x!=y)
			{
		 		f[y]=x;
			}
		 }
	}
	for(int i=1;i<=q;++i)
	{
		if(c[i])
		{
			flag=1;
			if(find(a[i])==find(b[i]))
			{
				puts("No");return 0;
			}
		}
	}
	int c=0;
	for(int i=1;i<=n;++i)
	{
		if(find(i)==i)
		{
	     c++;
		}
	}
	if(!flag)mi=c-1;
	else mi=max(3,c);
	ma=c*(c-1)/2;
	m-=(n-c);
	if(m>=mi&&m<=ma)puts("Yes");
	else puts("No");
}