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>
#define no return puts("No")&0;
#define yes return puts("Yes")&0;
using namespace std;
const int maxn=100005;
long long m;
int n,Q,cnt,a[maxn],b[maxn],c[maxn],fa[maxn];
inline char nc(){
	static char buf[100000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline long long read(){
	long long ret=0;bool f=0;char ch=nc();
	while(ch>'9'||ch<'0') f^=ch=='-',ch=nc();
	while(ch<='9'&&ch>='0') ret=ret*10+ch-'0',ch=nc();
	return f?-ret:ret;
}
int getfa(int x){return fa[x]==x?x:fa[x]=getfa(fa[x]);}
int main(){
	n=read(),m=read(),Q=read();
	for (int i=1;i<=n;i++) fa[i]=i;
	for (int i=1;i<=Q;i++){
		a[i]=read()+1,b[i]=read()+1,c[i]=read();
		if(m==n-1&&c[i]) no
		if(!c[i]){
			int fx=getfa(a[i]),fy=getfa(b[i]);
			if(fx^fy) fa[fx]=fy;
		}
	}
	if(m==n-1) yes
	for (int i=1;i<=Q;i++)
	if(c[i]&&getfa(a[i])==getfa(b[i])) no
	for (int i=1;i<=n;i++)
	cnt+=getfa(i)==i;
	if(cnt<=2) no
	if(m-n<=1ll*cnt*(cnt-3)>>1) yes
	else no
}