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 <algorithm>
using namespace std;
#define N 100005
#define ll long long
struct node{int u, v, c;}E[N];
ll m, sum;
int bel[N], n, q;
int Getbel(int x)
{
	if (!bel[x])
		return x;
	return bel[x] = Getbel(bel[x]);
}
int main()
{
	scanf("%d%lld%d", &n, &m, &q);
	for (int i = 1; i <= q; i++)
	{
		scanf("%d%d%d", &E[i].u, &E[i].v, &E[i].c);
		++E[i].u, ++E[i].v;
		if (!E[i].c)
		{
			int u = Getbel(E[i].u);
			int v = Getbel(E[i].v);
			if (u != v)
				bel[u] = v;
		}
		if (E[i].c && m == n - 1)
		{
			puts("No");
			return 0;
		}
	}
	if (m == n - 1)
	{
		puts("Yes");
		return 0;
	}
	for (int i = 1; i <= q; i++)
	{
		if (E[i].c)
		{
			int u = Getbel(E[i].u);
			int v = Getbel(E[i].v);
			if (u == v)
			{
				puts("No");
				return 0;
			}
		}
	}
	for (int i = 1; i <= n; i++)
		sum += !bel[i];
	sum = n - sum + (1ll * sum * (sum - 1) / 2);
	if (m <= sum)
		puts("Yes");
	else
		puts("No");
}