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<algorithm>
using namespace std;
typedef long long ll;
ll fx,fy,num=0,siz=0,r,n,x,y,lim,l,z,i,m,q,c[2][100005],tot=0,dsu[100005],vis[100005];
ll getf(ll x){return dsu[x]==x?x:dsu[x]=getf(dsu[x]);}
ll wan(ll x){return x*(x-1LL)/2LL;}
int main()
{
	scanf("%lld%lld%lld",&n,&m,&q);
	for(i=1;i<=n;i++)dsu[i]=i;
	for(i=1;i<=q;i++)
	{
		scanf("%lld%lld%lld",&x,&y,&z);
		x++;y++;
		if(z==0)
		{
			vis[x]=vis[y]=1;
			fx=getf(x);
			fy=getf(y);
			if(fx!=fy)dsu[fx]=fy;
		}
		else
		{
			tot++;
			c[0][tot]=x;
			c[1][tot]=y;
		}
	}
	for(i=1;i<=tot;i++)
	{
		fx=getf(c[0][i]);fy=getf(c[1][i]);
		if(fx==fy)
		{
			printf("No\n");
			return 0;
		}
	}
	for(i=1;i<=n;i++)
	{
		if(vis[i])siz++;
		if(vis[i]&&getf(i)==i)num++;
	}
	if(num>=3)lim=siz-num+wan(num+n-siz);
	if(num==2)
	{
		if(siz==n&&tot>0)
		{
			printf("No\n");
			return 0;
		}
		lim=siz-num+wan(num+n-siz);
	}
	if(num==1)
	{
		if(siz==n-1&&tot>0)
		{
			printf("No\n");
			return 0;
		}
		lim=siz-num+wan(num+n-siz);
	}
	if(num==0)lim=siz-num+wan(n+num-siz);
	if(tot==0)l=n-1;
	else l=n;
	if(m<=lim&&m>=l)printf("Yes\n");
	else printf("No\n");
	return 0;
}