#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;

int gi() {
	int x=0,o=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,q,fa[N],x[N],y[N],tt;
long long m;

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

void No() {
	cout<<"No";exit(0);
}

int main() {
	cin>>n>>m>>q;
	for(int i=1;i<=n;i++) fa[i]=i;
	int blo=n;
	for(int i=1;i<=q;i++) {
		int a,b,c;cin>>a>>b>>c;
		++a,++b;
		if(!c) {
			if(getf(a)!=getf(b)) fa[getf(a)]=getf(b),--blo;
		}
		else x[++tt]=a,y[tt]=b;
	}
	for(int i=1;i<=tt;i++)
		if(m<n||getf(x[i])==getf(y[i])) No();
	if(m>n-blo+1ll*blo*(blo-1)/2) No();
	else cout<<"Yes";
	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;
const int N=1e5+10;

int gi() {
	int x=0,o=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,q,fa[N],x[N],y[N],tt;
long long m;

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

void No() {
	cout<<"No";exit(0);
}

int main() {
	cin>>n>>m>>q;
	for(int i=1;i<=n;i++) fa[i]=i;
	int blo=n;
	for(int i=1;i<=q;i++) {
		int a,b,c;cin>>a>>b>>c;
		++a,++b;
		if(!c) {
			if(getf(a)!=getf(b)) fa[getf(a)]=getf(b),--blo;
		}
		else x[++tt]=a,y[tt]=b;
	}
	for(int i=1;i<=tt;i++)
		if(m<n||getf(x[i])==getf(y[i])) No();
	if(m>n-blo+1ll*blo*(blo-1)/2) No();
	else cout<<"Yes";
	return 0;
}

```