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;
struct union_find{
	static const int DEFAULT = 200200;
	vector<int> parent;
	int N, num_comps;
	union_find(int N = DEFAULT) : N(N) {
		clear();
	}
	void clear(){
		num_comps = N;
		parent.assign(N, -1);
	}
	int find(int u){
		return (parent[u] < 0) ? u : parent[u] = find(parent[u]);
	}
	bool is_connected(int u, int v){
		return find(u) == find(v);
	}
	bool join(int u, int v){
		u = find(u);
		v = find(v);
		if (u != v){
			num_comps--;
			if (parent[u] > parent[v])
				swap(u, v);
			parent[u] += parent[v];
			parent[v] = u;
		}
		return u != v;
	}
	int size(int u){
		return -parent[find(u)];
	}
	int comps(){
		return num_comps;
	}
};
int main(){
	int n, q; long long m; cin >> n >> m >> q;
	vector<pair<int,int> > bad;
	union_find uf(n);
	for(int i = 0; i < q; i++){
		int u, v, w; cin >> u >> v >> w;
		if(w) bad.push_back({u, v});
		else uf.join(u, v);
	}
	for(auto e : bad) if(uf.is_connected(e.first, e.second)) return !printf("No\n");
	if(m == n - 1){
		if(bad.empty()) printf("Yes\n");
		else printf("No\n");
		return 0;
	}
	int cc = uf.comps();
	if(m <= n - cc + 1ll * cc * (cc - 1) / 2) printf("Yes\n");
	else printf("No\n");
	return 0;
}