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 = 110000;

struct Edge {
	int u, v, t;
	Edge(int u = 0, int v = 0, int t = 0) : u(u), v(v), t(t) {}
} E[N];

int root[N];

int find_root(int u) {
	if (u == root[u]) return root[u];
	return root[u] = find_root(root[u]);
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	int n, q;
	long long m; cin >> n >> m >> q;
	for (int i = 0; i < q; i++) cin >> E[i].u >> E[i].v >> E[i].t, E[i].u++, E[i].v++;
	for (int i = 1; i <= n; i++) root[i] = i;
	for (int i = 0; i < q; i++) if (E[i].t == 0) {
		int a = find_root(E[i].u), b = find_root(E[i].v);
		if (a == b) continue;
		root[a] = b, m--;
	}
	int cnt = 0;
	for (int i = 0; i < q; i++) if (E[i].t == 1) {
		cnt++;
		int a = find_root(E[i].u), b = find_root(E[i].v);
		if (a == b) {
			cout << "No\n";
			return 0;
		}
	}
	int now = 0;
	for (int i = 1; i <= n; i++) now += root[i] == i;
	if ((cnt && now <= 2) || (m < now - (cnt == 0) || m > (long long)now * (now - 1) / 2)) {
		cout << "No\n";
		return 0;
	}
	cout << "Yes\n";
	return 0;
}