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>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>

using namespace std;

int fa[100050];

struct fuck
{
	int u,v,w;
	bool operator < (const fuck &a) const
	{
		return w < a.w;
	}
}Q[100050];

int n;
long long m;

int find_fa(int pos)
{
	return fa[pos] == pos ? pos : fa[pos] = find_fa(fa[pos]);
}

int main()
{
	int q;
	scanf("%d%lld%d",&n,&m,&q);
	for(int i = 1;i <= q; ++ i)
		scanf("%d%d%d",&Q[i].u,&Q[i].v,&Q[i].w),Q[i].u ++,Q[i].v ++;
	sort(Q + 1,Q + 1 + q);
	for(int i = 1;i <= n; ++ i) fa[i] = i;
	int fuck = 1;
	for(int i = 1;i <= q; ++ i)
	{
		if(!Q[i].w)
		{
			int u = find_fa(Q[i].u),v = find_fa(Q[i].v);
			fa[u] = v;
		}
		else
		{
			fuck = 0;
			if(find_fa(Q[i].u) == find_fa(Q[i].v))
			{
				printf("No\n");
				return 0;
			}
		}
	}
	int cnt = 0;
	for(int i = 1;i <= n; ++ i) if(fa[i] == i) cnt ++;
	long long cmax = n - cnt + cnt * 1ll * (cnt - 1) / 2;
	if(cmax >= m && n - fuck <= m) printf("Yes\n");
	else printf("No\n");
	return 0;
}