#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e5+2;
pair<int,pair<int,int> > query[N];
int head[N];
int findd(int x){
	if(head[x]<0){
		return x;
	}
	return head[x]=findd(head[x]);
}
void unionn(int x,int y){
	x=findd(x),y=findd(y);
	if(x==y){
		return;
	}
	head[x]+=head[y];
	head[y]=x;
}
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n,i,j,k,l,m,q;
	cin>>n>>m>>q;
	for(i=1;i<=q;i++){
		cin>>j>>k>>l;
		query[i]={l,{j,k}};
	}
	sort(query+1,query+1+q);
	for(i=1;i<=n;i++){
		head[i]=-1;
	}
	for(i=1;i<=q;i++){
		j=query[i].second.first+1,k=query[i].second.second+1;
		if(query[i].first==0){
			unionn(j,k);
		}
		else{
			if(m==n-1){
				cout<<"No";
				return 0;
			}
			if(findd(j)==findd(k)){
				cout<<"No";
				return 0;
			}
		}
	}
	if(m==n-1){
		cout<<"Yes";
		return 0;
	}
	j=0;
	for(i=1;i<=n;i++){
		if(i==findd(i)){
			j++;
		}
	}
	if(j==2||m>j*(j-3)/2+n){
		cout<<"No";
		return 0;
	}
	cout<<"Yes";
}
### Prompt

Construct a CPP code solution to the problem outlined:

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

### Response

```cpp
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e5+2;
pair<int,pair<int,int> > query[N];
int head[N];
int findd(int x){
	if(head[x]<0){
		return x;
	}
	return head[x]=findd(head[x]);
}
void unionn(int x,int y){
	x=findd(x),y=findd(y);
	if(x==y){
		return;
	}
	head[x]+=head[y];
	head[y]=x;
}
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n,i,j,k,l,m,q;
	cin>>n>>m>>q;
	for(i=1;i<=q;i++){
		cin>>j>>k>>l;
		query[i]={l,{j,k}};
	}
	sort(query+1,query+1+q);
	for(i=1;i<=n;i++){
		head[i]=-1;
	}
	for(i=1;i<=q;i++){
		j=query[i].second.first+1,k=query[i].second.second+1;
		if(query[i].first==0){
			unionn(j,k);
		}
		else{
			if(m==n-1){
				cout<<"No";
				return 0;
			}
			if(findd(j)==findd(k)){
				cout<<"No";
				return 0;
			}
		}
	}
	if(m==n-1){
		cout<<"Yes";
		return 0;
	}
	j=0;
	for(i=1;i<=n;i++){
		if(i==findd(i)){
			j++;
		}
	}
	if(j==2||m>j*(j-3)/2+n){
		cout<<"No";
		return 0;
	}
	cout<<"Yes";
}
```