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;

int n,q,x,y,v,hd[100010],cnt,fa[100010];
struct node{int to,next,val; bool bo;}e[200010];
long long m;

int ask(int x) {return fa[x]==x?x:fa[x]=ask(fa[x]);}

void addedge(int x,int y,int v)
{
	e[++cnt]=(node){y,hd[x],v,0},hd[x]=cnt;
	e[++cnt]=(node){x,hd[y],v,0},hd[y]=cnt;
}

int main()
{
	scanf("%d%lld%d",&n,&m,&q);
	for (int i=1; i<=q; i++) scanf("%d%d%d",&x,&y,&v),addedge(x+1,y+1,v);
	for (int i=1; i<=n; i++) fa[i]=i;
	int cnt0=0,cnt1=0;
	for (int i=1; i<=q; i++) if (!e[i*2].val)
	{
		if (ask(e[i*2].to)==ask(e[i*2-1].to)) continue;
		fa[ask(e[i*2].to)]=ask(e[i*2-1].to),cnt0++;
	}
	for (int i=1; i<=q; i++) if (e[i*2].val)
	{
		cnt1++;
		if (ask(e[i*2].to)==ask(e[i*2-1].to)) return puts("No"),0;
	}
	cnt=n-cnt0,m-=cnt0;
	if (m>1ll*cnt*(cnt-1)/2) return puts("No"),0;
	if (m<cnt-1+(cnt1>0)) return puts("No"),0;
	puts("Yes");
	return 0;
}