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 <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <fstream>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <map>
#include <set>

using namespace std;

typedef long long ll;

const int MAXN = 200005;

int n,Q;
int fa[MAXN];
int a[MAXN];
int b[MAXN];
int c[MAXN];
int sz[MAXN];

ll m;

int getroot(int p)
{
	return p == fa[p] ? p : fa[p] = getroot(fa[p]);
}

int main()
{
	scanf("%d%lld%d",&n,&m,&Q);
	for (int i = 1;i <= n;i++)
		fa[i] = i;
	for (int i = 1;i <= Q;i++)
	{
		scanf("%d%d%d",&a[i],&b[i],&c[i]);
		a[i]++,b[i]++;
		if (c[i] == 0)
		{
			if (getroot(a[i]) != getroot(b[i]))
				m--;
			fa[getroot(a[i])] = getroot(b[i]);
		}
	}
	bool any1 = 0;
	for (int i = 1;i <= Q;i++)
		if (c[i] == 1)
		{
			any1 = 1;
			a[i] = getroot(a[i]);
			b[i] = getroot(b[i]);
			if (a[i] == b[i])
			{
				puts("No");
				return 0;
			}
		}
	for (int i = 1;i <= n;i++)
		sz[getroot(i)]++;
	ll t = 0;
	for (int i = 1;i <= n;i++)
		t += (sz[i] > 0);
	if (m > t * (t - 1) / 2 || (t == 2 && any1) || m < t - 1 + any1)
	{
		puts("No");
		return 0;
	}
	puts("Yes");
	return 0;
}