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;
#define mod 998244353
#define MOD 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3fll
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
const int maxn=1e5+10;
int fa[maxn];
inline int getfa(int x){return fa[x]==x?x:fa[x]=getfa(fa[x]);}
int u[maxn],v[maxn],tp[maxn];
inline bool Union(int x,int y)
{
	x=getfa(x),y=getfa(y);
	if(x==y)return 0;
	fa[x]=y;
	return 1;
}
int main()
{
	int n,q;
	ll m;
	bool f=0;
	scanf("%d%lld%d",&n,&m,&q);
	for(int i=0;i<n;i++)
		fa[i]=i;
	for(int i=0;i<q;i++)
	{
		scanf("%d%d%d",&u[i],&v[i],&tp[i]);
		if(!tp[i])Union(u[i],v[i]);
		if(tp[i])f=1;
	}
	if(m==n-1&&f)
	{
		printf("No\n");
		return 0;
	}
	for(int i=0;i<q;i++)
	{
		if(tp[i]&&getfa(u[i])==getfa(v[i]))
		{
			printf("No\n");
			return 0;
		}
	}
	int cnt=0;
	for(int i=0;i<n;i++)
		if(getfa(i)==i)cnt++;//cout<<i<<" "<<getfa(i)<<endl;
	if(f==1&&cnt<=2)
	{
		printf("No\n");
		return 0;
	}
	ll mx=cnt*1ll*(cnt-1)/2ll+(n-cnt);
/*	set<int> st;
	st.clear();
	for(int i=0;i<q;i++)
	{
		if(tp[i])
		{
			st.insert(getfa(u[i]));
			st.insert(getfa(v[i]));
		}
	}*/
	if(mx>=m)
	{
		printf("Yes\n");
	}
	else
	{
		printf("No\n");
	}
	return 0;
}