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 ll long long

struct node { int x,y,z; }a[100010];

int fa[100010],n,m,q;

inline int rd()
{
	int x=0;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar());
	for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x;
}

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

int main()
{
	n=rd();m=rd();q=rd();
	if (m==n-1)
	{
		for (int i=1;i<=q;i++)
		{
			rd();rd();int z=rd();
			if (z==1) { puts("No");return 0; }
		}
		puts("Yes");
		return 0;
	}
	for (int i=1;i<=n;i++) fa[i]=i;
	for (int i=1;i<=q;i++) a[i].x=rd()+1,a[i].y=rd()+1,a[i].z=rd();
	for (int i=1;i<=q;i++) if (a[i].z==0) fa[find(a[i].x)]=find(a[i].y);
	int cnt=0,cnt1=0;
	for (int i=1;i<=n;i++) if (find(i)==i) cnt++;
	for (int i=1;i<=q;i++) if (a[i].z==1&&find(a[i].x)==find(a[i].y)) { puts("No");return 0; }
	for (int i=1;i<=q;i++) if (a[i].z==1) cnt1++;
	if (cnt1&&cnt==2) { puts("No");return 0; }
	if (cnt==1) { puts("No");return 0; }
	ll mn=(cnt==2)?1:cnt,mx=(ll)cnt*(cnt-1)/2;
	if (n-cnt+mn<=m&&n-cnt+mx>=m) puts("Yes");
	else puts("No");
	return 0;
}