#include<bits/stdc++.h>
const int N=1e5+5;
typedef long long ll;
int n,q,a[N],b[N],c[N],i,g[N],sz[N],C,CC;
ll m;
int gfa(int x){return g[x]==x?x:g[x]=gfa(g[x]);}
int main(){
	scanf("%d%lld%d",&n,&m,&q);
	for(i=1;i<=n;++i)g[i]=i;
	for(i=1;i<=q;++i)scanf("%d%d%d",a+i,b+i,c+i),++a[i],++b[i],c[i]==0?g[gfa(a[i])]=gfa(b[i]):0;
	for(i=1;i<=q;++i)if(c[i]==1){if(gfa(a[i])==gfa(b[i]))return puts("No"),0;C=1;}
	for(i=1;i<=n;++i)++sz[gfa(i)];
	for(i=1;i<=n;++i)if(gfa(i)==i)++CC;
	puts(n-1+C<=m && m<=1ll*CC*(CC-1)/2+n-CC && (!C || CC>2)?"Yes":"No");
	return 0;
}
### 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>
const int N=1e5+5;
typedef long long ll;
int n,q,a[N],b[N],c[N],i,g[N],sz[N],C,CC;
ll m;
int gfa(int x){return g[x]==x?x:g[x]=gfa(g[x]);}
int main(){
	scanf("%d%lld%d",&n,&m,&q);
	for(i=1;i<=n;++i)g[i]=i;
	for(i=1;i<=q;++i)scanf("%d%d%d",a+i,b+i,c+i),++a[i],++b[i],c[i]==0?g[gfa(a[i])]=gfa(b[i]):0;
	for(i=1;i<=q;++i)if(c[i]==1){if(gfa(a[i])==gfa(b[i]))return puts("No"),0;C=1;}
	for(i=1;i<=n;++i)++sz[gfa(i)];
	for(i=1;i<=n;++i)if(gfa(i)==i)++CC;
	puts(n-1+C<=m && m<=1ll*CC*(CC-1)/2+n-CC && (!C || CC>2)?"Yes":"No");
	return 0;
}
```