#include<bits/stdc++.h>

using namespace std;

#define N 300000

int n,m,p,cnt,tmp,fa[N],a[N],b[N],c[N];
bool vis[N];

int gfa(int x){return fa[x]==x?x:fa[x]=gfa(fa[x]);}

void ins(int x){
	if (!vis[x=gfa(x)]){vis[x]=1; ++tmp;}
}

int main(){
	scanf("%d%d%d",&n,&m,&p);
	cnt=n;
	for (int i=0;i<n;++i) fa[i]=i;
	for (int i=1;i<=p;++i){
		scanf("%d%d%d",a+i,b+i,c+i);
		if (c[i]==0){
			if (gfa(a[i])!=gfa(b[i])){
				--cnt;
				fa[gfa(a[i])]=gfa(b[i]);
			}
		}
	}
	for (int i=1;i<=p;++i)
		if (c[i]==1){
			if (gfa(a[i])==gfa(b[i])){puts("No"); return 0;}
			ins(a[i]); ins(b[i]);
		}
	m=m-n+cnt;
	if (m>(long long)cnt*(cnt-1)/2){
		puts("No"); return 0;
	}
	if (tmp) m-=max(3,tmp);
	puts(m>=0?"Yes":"No");
	
	return 0;
}
			
### Prompt

Please formulate a Cpp solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>

using namespace std;

#define N 300000

int n,m,p,cnt,tmp,fa[N],a[N],b[N],c[N];
bool vis[N];

int gfa(int x){return fa[x]==x?x:fa[x]=gfa(fa[x]);}

void ins(int x){
	if (!vis[x=gfa(x)]){vis[x]=1; ++tmp;}
}

int main(){
	scanf("%d%d%d",&n,&m,&p);
	cnt=n;
	for (int i=0;i<n;++i) fa[i]=i;
	for (int i=1;i<=p;++i){
		scanf("%d%d%d",a+i,b+i,c+i);
		if (c[i]==0){
			if (gfa(a[i])!=gfa(b[i])){
				--cnt;
				fa[gfa(a[i])]=gfa(b[i]);
			}
		}
	}
	for (int i=1;i<=p;++i)
		if (c[i]==1){
			if (gfa(a[i])==gfa(b[i])){puts("No"); return 0;}
			ins(a[i]); ins(b[i]);
		}
	m=m-n+cnt;
	if (m>(long long)cnt*(cnt-1)/2){
		puts("No"); return 0;
	}
	if (tmp) m-=max(3,tmp);
	puts(m>=0?"Yes":"No");
	
	return 0;
}
			
```