#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
const int N=1e5+10;
ll n,m,q,dlim,ulim;
int z1[N],z2[N],h1[N],h2[N];
int fa[N],t1,t2,cnt[N];
int gf(int x){return fa[x]==x?x:fa[x]=gf(fa[x]);}
void write(int p){
	puts(p?"Yes":"No");
}
int main(){
	cin>>n>>m>>q;
	for (int i=1;i<=q;i++){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		if (c==0){
			z1[++t1]=a+1;
			z2[t1]=b+1;
		}
		else {
			h1[++t2]=a+1;
			h2[t2]=b+1;
		}
	}
	for (int i=1;i<=n;i++)fa[i]=i;
	dlim=n; 
	for (int i=1;i<=t1;i++){
		if (gf(z1[i])!=gf(z2[i]))
		fa[gf(z1[i])]=gf(z2[i]),dlim--;
	}
	for (int i=1;i<=t2;i++){
		if (gf(h1[i])==gf(h2[i])){
			printf("No\n");
			return 0;
		}
	}
	ulim=(n-dlim)+dlim*(dlim-1)/2;
	if (dlim==2){
		if (t2)write(0);
		else write(m==n-1);
		return 0;
	}
	if (t2)dlim=n;
	else dlim=n-1;
	write(dlim<=m&&m<=ulim);
	
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
const int N=1e5+10;
ll n,m,q,dlim,ulim;
int z1[N],z2[N],h1[N],h2[N];
int fa[N],t1,t2,cnt[N];
int gf(int x){return fa[x]==x?x:fa[x]=gf(fa[x]);}
void write(int p){
	puts(p?"Yes":"No");
}
int main(){
	cin>>n>>m>>q;
	for (int i=1;i<=q;i++){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		if (c==0){
			z1[++t1]=a+1;
			z2[t1]=b+1;
		}
		else {
			h1[++t2]=a+1;
			h2[t2]=b+1;
		}
	}
	for (int i=1;i<=n;i++)fa[i]=i;
	dlim=n; 
	for (int i=1;i<=t1;i++){
		if (gf(z1[i])!=gf(z2[i]))
		fa[gf(z1[i])]=gf(z2[i]),dlim--;
	}
	for (int i=1;i<=t2;i++){
		if (gf(h1[i])==gf(h2[i])){
			printf("No\n");
			return 0;
		}
	}
	ulim=(n-dlim)+dlim*(dlim-1)/2;
	if (dlim==2){
		if (t2)write(0);
		else write(m==n-1);
		return 0;
	}
	if (t2)dlim=n;
	else dlim=n-1;
	write(dlim<=m&&m<=ulim);
	
}
```