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>
#define X first
#define Y second
#define LL long long
#define MP make_pair
#define PB push_back
#define SZ(a) (LL)(a.size())
const LL INF = 1000 * 1000 * 1000 + 47;
const LL LINF = INF * (LL)INF;
using namespace std;
const LL MAX = 100005;
const LL MAGIC = 50;
int p[MAX];
LL comps;
LL m;
int n , q;
int u[MAX] , v[MAX] , type[MAX];
void pzdc()
{
	cout << "No" << endl;
	exit(0);
}
int find(int u)
{
	return p[u] = u == p[u] ? u : find(p[u]);
}
void unite(int u , int v)
{
	u = find(u); v = find(v);
	if(u == v)
		return;
	comps--;
	p[v] = u;
}
int main() 
{
	cin >> n >> m >> q;
	for(int i = 0; i < n; i++)
		p[i] = i;
	comps = n;
	bool was1 = false;
	for(int i = 0; i < q; i++)
	{
		cin >> u[i] >> v[i] >> type[i];
		was1 |= type[i];
		if(type[i] == 0)
			unite(u[i] , v[i]);
	}
	for(int i = 0; i < q; i++)
		if(type[i] == 1)
			if(find(u[i]) == find(v[i]))
				pzdc();
	LL lBound = n - 1 + was1 , rBound = n - comps + comps * (comps - 1) / 2;
	if(m < lBound || m > rBound)
		pzdc();
	
	cout << "Yes" << endl;
	return 0;
}