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;
int n,q;
long long m;
vector<int>adj[100001];
int u[100001],v[100001],w[100001];
bool tree;
int cc;
int par[100001];
int find(int x){
	if(par[x]!=x) par[x]=find(par[x]);
	return par[x];
}
void join(int x,int y){
	x=find(x);y=find(y);
	cc-=(x!=y);
	par[x]=y;
}
int main(){
	ios::sync_with_stdio(false);
	cin >> n >> m >> q;
	tree=m==n-1;cc=n;
	for(int i=1; i<=n ;i++){
		par[i]=i;
	}
	for(int i=1; i<=q ;i++){
		cin >> u[i] >> v[i] >> w[i];
		if(w[i]==0) join(u[i],v[i]);
		tree&=(w[i]!=1);
	}
	for(int i=1; i<=q ;i++){
		if(!w[i]) continue;
		if(find(u[i])==find(v[i])) return cout << "No\n",0;
	}
	if(tree) return cout << "Yes\n",0;
	if(cc<=2) return cout << "No\n",0;
	if(m>=n && m<=n+1LL*cc*(cc-3)/2) cout << "Yes\n";
	else cout << "No\n";
}