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>
#define ll long long
using namespace std;
struct aaa{
	ll x,y,op;
}a[200010];
ll t1,n,m,sz,flag,fa[200010];
ll find(ll x){
	if(x==fa[x])return x;
	return fa[x]=find(fa[x]);
}
int main(){
	ll i,x,y,mn,mx;
	scanf("%lld%lld%lld",&n,&m,&t1);
	sz=n;
	for(i=1;i<=n;i++)fa[i]=i;
	for(i=1;i<=t1;i++){
		scanf("%lld%lld%lld",&a[i].x,&a[i].y,&a[i].op);
		a[i].x++;a[i].y++;
		if(a[i].op==0){
			x=find(a[i].x);y=find(a[i].y);
			if(x!=y){
				fa[x]=y;sz--;
			}
		}
	}
	for(i=1;i<=t1;i++)if(a[i].op==1&&find(a[i].x)==find(a[i].y))return puts("No"),0;
	for(i=1;i<=t1;i++)if(a[i].op)flag=1;
	if(!flag)mn=sz-1;
	 else mn=sz;
	mx=sz*(sz-1)/2;
	if(flag&&sz<=2)return puts("No"),0;
	m=m-(n-sz);
	if(mn<=m&&m<=mx)puts("Yes");
	 else puts("No");
}