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<iostream>
#include<cstdio>
#include<vector>
#include<tuple>
using namespace std;
const int N=100005;
int n;
long long m;
int Q;
int fa[N];
int getf(int v)
{
	if(v==fa[v]) return v;
	fa[v]=getf(fa[v]);
	return fa[v];
}
bool merge(int u,int v)
{
	int f1=getf(u),f2=getf(v);
	if(f1!=f2)
	{
		fa[f2]=f1;
		return true;
	}
	else return false;
}
int main()
{
	scanf("%d%lld%d",&n,&m,&Q);
	for(int i=1;i<=n;i++)
		fa[i]=i;
	vector<tuple<int,int,int> >G;
	int num=0;
	for(int i=1;i<=Q;i++)
	{
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		u++,v++;
		if(w==0) num+=merge(u,v);
		else if(w==1) G.push_back(make_tuple(u,v,w));
	}
	for(auto [u,v,w]:G)
		if(getf(u)==getf(v))
		{
			printf("No");
			return 0;
		}
	int cnt=0;
	for(int i=1;i<=n;i++)
		if(getf(i)==i) cnt++;
	if(cnt==2&&!G.empty())
	{
		printf("No");
		return 0;
	}
	long long l=num+cnt,r=num+1LL*cnt*(cnt-1)/2;
	if(cnt==1) l=num,r=num;
	if(G.empty()) l=num-1;
	if(l<=m&&m<=r) printf("Yes");
	else printf("No");
	return 0;
}