#include<cstdio>
const int MaxN=100010,MaxQ=100010;
int N,Q,A[MaxQ],B[MaxQ],C[MaxQ],fa[MaxN];
long long M;
int find(int i){return fa[i]==i?i:fa[i]=find(fa[i]);}
int main(){
	scanf("%d%lld%d",&N,&M,&Q);
	for(int i=0;i<N;i++)fa[i]=i;
	int blo=N;
	for(int i=0;i<Q;i++){
		scanf("%d%d%d",A+i,B+i,C+i);
		if(C[i]==0&&find(A[i])!=find(B[i]))fa[find(A[i])]=find(B[i]),blo--;
	}
	for(int i=0;i<Q;i++)if(C[i]==1){
		if(M<N||find(A[i])==find(B[i])||blo==2)return puts("No"),0;
	}
	if(M-(N-blo)>1ll*blo*(blo-1)/2)puts("No");
	else puts("Yes");
}
### Prompt

Your task is to create 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<cstdio>
const int MaxN=100010,MaxQ=100010;
int N,Q,A[MaxQ],B[MaxQ],C[MaxQ],fa[MaxN];
long long M;
int find(int i){return fa[i]==i?i:fa[i]=find(fa[i]);}
int main(){
	scanf("%d%lld%d",&N,&M,&Q);
	for(int i=0;i<N;i++)fa[i]=i;
	int blo=N;
	for(int i=0;i<Q;i++){
		scanf("%d%d%d",A+i,B+i,C+i);
		if(C[i]==0&&find(A[i])!=find(B[i]))fa[find(A[i])]=find(B[i]),blo--;
	}
	for(int i=0;i<Q;i++)if(C[i]==1){
		if(M<N||find(A[i])==find(B[i])||blo==2)return puts("No"),0;
	}
	if(M-(N-blo)>1ll*blo*(blo-1)/2)puts("No");
	else puts("Yes");
}
```