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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define ll long long
#define fr(i,x,y) for(int i=(x);i<=(y);i++)
#define rf(i,x,y) for(int i=(x);i>=(y);i--)
#define frl(i,x,y) for(int i=(x);i<(y);i++)
using namespace std;
const int N=200002;
int n,Q;
ll m;
struct data{
	int x,y,w;
	bool operator < (const data &q)const{ return w<q.w; }
}a[N];
int f[N];

template<class T> void read(T &x){
	char ch=getchar();x=0;
	for(;ch<'0'||ch>'9';ch=getchar());
	for(;ch>='0'&&ch<='9';ch=getchar()) x=(x<<3)+(x<<1)+ch-'0';
}

int gf(int x){ return f[x]==x?x:f[x]=gf(f[x]); }

int main(){
	read(n);read(m);read(Q);
	fr(i,1,Q) read(a[i].x),read(a[i].y),read(a[i].w);
	fr(i,1,Q) a[i].x++,a[i].y++;
	sort(a+1,a+1+Q);
	fr(i,1,n) f[i]=i;
	int flag=0;
	fr(i,1,Q){
		if (a[i].w){
			flag=i;
			break;
		}
		int x=a[i].x,y=a[i].y;
		f[gf(x)]=gf(y);
	}
	if (flag) fr(i,flag,Q) if (gf(a[i].x)==gf(a[i].y)) return puts("No"),0;
	int cnt=0;
	fr(i,1,n) if (gf(i)==i) cnt++;
	if (!flag){
		if (m>1LL*cnt*(cnt-1)/2+n-cnt) puts("No");
		 else puts("Yes");
	} else{
		if (cnt<=2||m<n) puts("No");
		 else if (m>1LL*cnt*(cnt-1)/2+n-cnt) puts("No");
		  else puts("Yes");
	}
	return 0;
}