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<stdio.h>
#define ll long long
using namespace std;
int fa[200005];
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
int l[200005],r[200005];
ll n,m,k,c,cnt;
int main(){
	scanf("%lld%lld%lld",&n,&m,&k);
	for(int i=1;i<=n;i++) fa[i]=i;
	cnt=0;
	while(k--){
		ll u,v,opt;scanf("%lld%lld%lld",&u,&v,&opt);u++;v++;
		if(!opt) fa[find(u)]=find(v);
		else cnt++,l[cnt]=u,r[cnt]=v;
	}
	for(int i=1;i<=cnt;i++) if(find(l[i])==find(r[i])) return puts("No"),0;
	for(int i=1;i<=n;i++) if(find(i)==i) c++;
	ll L=cnt?c:c-1ll,R=c*(c-1ll)/2ll;
	return puts((L<=m-n+c&&m-n+c<=R)?"Yes":"No"),0;
}