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<cstdio>
#define RI register int
#define CI const int&
using namespace std;
const int N=100005;
int n,q,a[N],b[N],c[N],cur; long long m;
class UnionFindSet
{
	private:
		int fa[N];
	public:
		inline void init(void)
		{
			for (RI i=1;i<=n;++i) fa[i]=i;
		}
		inline int getfa(CI x)
		{
			return x!=fa[x]?fa[x]=getfa(fa[x]):x;
		}
		inline void Union(int x,int y)
		{
			x=getfa(x); y=getfa(y); if (x!=y) fa[x]=y;
		}
		inline bool identify(CI x,CI y)
		{
			return getfa(x)==getfa(y);
		}
}S;
int main()
{
	RI i; for (scanf("%d%lld%d",&n,&m,&q),S.init(),i=1;i<=q;++i)
	{
		scanf("%d%d%d",&a[i],&b[i],&c[i]); ++a[i]; ++b[i];
		if (!c[i]) S.Union(a[i],b[i]); if (m==n-1&&c[i]) return puts("No"),0;
	}
	if (m==n-1) return puts("Yes"),0; for (i=1;i<=q;++i)
	if (c[i]&&S.identify(a[i],b[i])) return puts("No"),0;
	for (i=1;i<=n;++i) cur+=S.getfa(i)==i;
	return puts(m<=n+(1LL*cur*(cur-3)>>1)?"Yes":"No"),0;
}