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 next Next
#define last Last
#define int long long
const int N=1e6+5;
int n,m,k,cnt,gs,fa[N];
struct node{
	int x,y,opt;
}q[N];
/*char buf[1<<21],*p1=buf,*p2=buf;
inline int gc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}*/
#define gc getchar
inline int read()
{
	int ret=0,f=0;char c=gc();
	while(!isdigit(c)){if(c=='-')f=1;c=gc();}
	while(isdigit(c)){ret=ret*10+c-48;c=gc();}
	if(f)return -ret;return ret;
}
int find(int x)
{
	if(fa[x]==x)return x;
	fa[x]=find(fa[x]);
	return fa[x];
}
signed main()
{
	n=read();m=read();k=read();cnt=n;
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int i=1;i<=k;i++)
	{
		q[i].x=read()+1,q[i].y=read()+1,q[i].opt=read();
		if(q[i].opt==0)
		{
			if(find(q[i].x)==find(q[i].y))continue;
			fa[find(q[i].x)]=find(q[i].y);
			cnt--;
		}
		else gs++;
	}
	if(m==n-1)
	{
		if(gs)puts("No");
		else puts("Yes");
		return 0;
	}
	for(int i=1;i<=k;i++)
	{
		if(q[i].opt==1)
		{
			if(find(q[i].x)==find(q[i].y)){puts("No");return 0;}
		}
	}
	if(m-(n-cnt)<=cnt*(cnt-1)/2)puts("Yes");
	else puts("No");
}