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;
const int N = 1e6 + 7, mod = 998244353;
#define debug(x) cerr << #x << " " << x << "\n";
#define int long long
int rep[N];
vector <pair<int, int> > kra;

void no() {
	cout << "No\n";
	exit(0);
}

void yes() {
	cout << "Yes\n";
	exit(0);
}

int re(int x) {
	if(rep[x] != x) rep[x] = re(rep[x]);
	return rep[x];
}

void uni(int a, int b) {
	rep[re(a)] = re(b);
}

int32_t main() {
	ios_base::sync_with_stdio(0);
	cin.tie();
	long long n, m, q;
	cin >> n >> m >> q;
	for(int i = 1; i <= n; i++) rep[i] = i;

	for(int i = 0; i < q; i++) {
		int a, b, c;
		cin >> a >> b >> c;
		a++;
		b++;
		if(c == 0) {
			uni(a, b);
		} else {
			kra.push_back(make_pair(a, b));
		}
	}

	if(m == n - 1) {
		if(kra.size() == 0) yes();
		else no();
	}

	for(auto x : kra) {
		if(re(x.first) == re(x.second)) no();
	}

	set <int> S;
	for(int i = 1; i <= n; i++) S.insert(re(i));
	long long k = S.size();

	if(m <= n - k + ((k * (k - 1)) / 2)) yes();
	no();
}