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;
typedef long long ll;
const int N=100000;
int n,q;
ll e;
vector<pair<int,int> > E[2];
int p[N],sz[N];
int find(int u){
	if(p[u]==u)
		return u;
	return p[u]=find(p[u]);
}
int main() {
	cin>>n>>e>>q;
	for(int i=0;i<n;++i){
		p[i]=i;
		sz[i]=1;
	}
	vector<bool> seen(n);
	for(int i=0;i<q;++i){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		E[c].push_back({a,b});
		if(c==0){
			seen[a]=seen[b]=true;
			a=find(a);
			b=find(b);
			if(a!=b){
				p[a]=b;
				sz[b]+=sz[a];
			}
		}
	}
	for(auto &e:E[1])
		if(find(e.first)==find(e.second)){
			puts("No");
			return 0;
		}
	ll need=0;
	int last=-1;
	int comp=0;
	for(int i=0;i<n;++i)
		if(p[i]==i){
			if(sz[i]==1){
				if(last==-1)
					last=i;
				int u=find(last);
				int v=find(i);
				if(u!=v){
					p[u]=v;
					sz[v]+=sz[u];
				}
			}else{
				need+=sz[i]-1;
				++comp;
			}
		}
	int s=0;
	if(last!=-1)
		s=sz[find(last)];
	bool res=false;
	comp+=s;
	if(need+comp-E[1].empty()<=e && e<=need+(ll)comp*(comp-1)/2)
		res=true;
	if(res)
		puts("Yes");
	else
		puts("No");
	return 0;
}