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<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define ll long long
#define inf 0x3f3f3f3f
#define mod 998244353
#define maxn 200010
inline ll read()
{
	ll x=0; char c=getchar(),f=1;
	for(;c<'0'||'9'<c;c=getchar())if(c=='-')f=-1;
	for(;'0'<=c&&c<='9';c=getchar())x=x*10+c-'0';
	return x*f;
}
inline void write(ll x)
{
	static char buf[20];
	int len=0;
	if(x<0)putchar('-'),x=-x;
	for(;x;x/=10)buf[len++]=x%10+'0';
	if(!len)putchar('0');
	else while(len)putchar(buf[--len]);
}
inline void writesp(ll x){write(x); putchar(' ');}
inline void writeln(ll x){write(x); putchar('\n');}
int fa[maxn];
int u[maxn],v[maxn];
ll n,m,q;
int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
int main()
{
	n=read(); m=read(); q=read();
	if(m==n-1){
		for(int i=1;i<=q;i++){
			int x=read(),y=read(),z=read();
			if(z){
				puts("No"); return 0;
			}
		}
		puts("Yes"); return 0;
	}
	for(int i=0;i<n;i++)
		fa[i]=i;
	int cnt=n,tot=0;
	for(int i=1;i<=q;i++){
		int x=read(),y=read(),z=read();
		if(!z){
			int fx=find(x),fy=find(y);
			if(fx!=fy)--cnt,fa[fx]=fy;
		}
		else u[++tot]=x,v[tot]=y;
	}
	for(int i=1;i<=tot;i++)
		if(find(u[i])==find(v[i])){
			puts("No"); return 0;
		}
	if(m<=n-cnt+(ll)cnt*(cnt-1)/2)puts("Yes");
	else puts("No");
	return 0;
}