#include<bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define mp make_pair
#define pb push_back
#define fst first
#define snd second
typedef long long ll;
typedef pair<int,int> pii;

const int maxn=100005;
int n,q;
ll m;
int a[maxn],b[maxn],c[maxn];
bool bad[maxn],ok;
namespace Dsu{
	int f[maxn];
	void init(){
		rep(i,n)f[i]=i;
	}
	int find(int x){
		return f[x]==x?x:f[x]=find(f[x]);
	}
};

int main(){
	scanf("%d%lld%d",&n,&m,&q);
	Dsu::init();
	rep(i,q){
		scanf("%d%d%d",&a[i],&b[i],&c[i]);
		if(c[i]==0){
			int u=Dsu::find(a[i]),v=Dsu::find(b[i]);
			Dsu::f[u]=v;
		}
	}
	ok=true;
	rep(i,q)if(c[i]==1){
		int u=Dsu::find(a[i]),v=Dsu::find(b[i]);
		if(u==v){
			puts("No");
			return 0;
		}
		bad[u]=bad[v]=true;
		ok=false;
	}
	int cnt=0,tot=0;
	rep(i,n)if(Dsu::find(i)==i){
		tot++;
		if(!bad[i])cnt++;
	}
	if(tot==2&&cnt==0){
		puts("No");
		return 0;
	}
	if(!ok&&m==n-1){
		puts("No");
		return 0;
	}
	if(m<=n-tot+1LL*tot*(tot-1)/2)puts("Yes");
	else puts("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>
using namespace std;

#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define mp make_pair
#define pb push_back
#define fst first
#define snd second
typedef long long ll;
typedef pair<int,int> pii;

const int maxn=100005;
int n,q;
ll m;
int a[maxn],b[maxn],c[maxn];
bool bad[maxn],ok;
namespace Dsu{
	int f[maxn];
	void init(){
		rep(i,n)f[i]=i;
	}
	int find(int x){
		return f[x]==x?x:f[x]=find(f[x]);
	}
};

int main(){
	scanf("%d%lld%d",&n,&m,&q);
	Dsu::init();
	rep(i,q){
		scanf("%d%d%d",&a[i],&b[i],&c[i]);
		if(c[i]==0){
			int u=Dsu::find(a[i]),v=Dsu::find(b[i]);
			Dsu::f[u]=v;
		}
	}
	ok=true;
	rep(i,q)if(c[i]==1){
		int u=Dsu::find(a[i]),v=Dsu::find(b[i]);
		if(u==v){
			puts("No");
			return 0;
		}
		bad[u]=bad[v]=true;
		ok=false;
	}
	int cnt=0,tot=0;
	rep(i,n)if(Dsu::find(i)==i){
		tot++;
		if(!bad[i])cnt++;
	}
	if(tot==2&&cnt==0){
		puts("No");
		return 0;
	}
	if(!ok&&m==n-1){
		puts("No");
		return 0;
	}
	if(m<=n-tot+1LL*tot*(tot-1)/2)puts("Yes");
	else puts("No");
	return 0;
}
```