#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
typedef long long LL;
const int N = 1e5+5;

int n,q,a[N],b[N],c[N],fa[N];
LL m;

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

int main(){
	scanf("%d%lld%d",&n,&m,&q);
	rep(i,1,n) fa[i] = i;
	rep(i,1,q){
		scanf("%d%d%d",a+i,b+i,c+i),++a[i],++b[i];
		if(c[i]==0){
			int x = find(a[i]), y = find(b[i]);
			fa[x] = y;
		}
	}
	if(m==n-1){
		printf(*max_element(c+1,c+q+1)==0 ? "Yes" : "No");
		return 0;
	}
	rep(i,1,q)if(c[i]==1 && find(a[i])==find(b[i]) ){
		printf("No"); return 0;
	}
	int k = 0;
	rep(i,1,n) k += (i==fa[i]);
	printf(m>=n && m<=n-k+1ll*k*(k-1)/2 ? "Yes" : "No");
	return 0;
}
### 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<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
typedef long long LL;
const int N = 1e5+5;

int n,q,a[N],b[N],c[N],fa[N];
LL m;

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

int main(){
	scanf("%d%lld%d",&n,&m,&q);
	rep(i,1,n) fa[i] = i;
	rep(i,1,q){
		scanf("%d%d%d",a+i,b+i,c+i),++a[i],++b[i];
		if(c[i]==0){
			int x = find(a[i]), y = find(b[i]);
			fa[x] = y;
		}
	}
	if(m==n-1){
		printf(*max_element(c+1,c+q+1)==0 ? "Yes" : "No");
		return 0;
	}
	rep(i,1,q)if(c[i]==1 && find(a[i])==find(b[i]) ){
		printf("No"); return 0;
	}
	int k = 0;
	rep(i,1,n) k += (i==fa[i]);
	printf(m>=n && m<=n-k+1ll*k*(k-1)/2 ? "Yes" : "No");
	return 0;
}
```