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>
using namespace std;
typedef long long ll;
const int N=200025,inf=0x3f3f3f3f;
int read(){
	int f=1,g=0;
	char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
int n,Q,cnt,need,fa[N],siz[N],h[N];
ll m;
vector< pair<int,int> > a;
int find(int x){return (x==fa[x]) ? x : fa[x]=find(fa[x]);}
void merge(int x,int y){
	x=find(x);y=find(y);
	if (x==y) return;
	if (siz[x]<siz[y]) swap(x,y);
	fa[y]=x;siz[x]+=siz[y];
}
int main(){
//	freopen("balance.in","r",stdin);
//	freopen("balance.out","w",stdout);
	scanf("%d %lld %d",&n,&m,&Q);
	for (int i=1;i<=n;i++) fa[i]=i,siz[i]=1;
	while (Q--){
		int u=read()+1,v=read()+1,typ=read();
		if (typ) a.push_back(make_pair(u,v));
		else merge(u,v);
	}
	for (int i=1;i<=n;i++) if (fa[i]==i) {cnt++;m-=siz[i]-1;}
	if (m<0) {puts("No");return 0;}
	for (auto t : a){
		int u=t.first,v=t.second;
		if (find(u)==find(v)) {puts("No");return 0;}
		else{
			if (!h[find(u)]) need++;
			if (!h[find(v)]) need++;
			h[find(u)]=h[find(v)]=1;
		}
	}
	if (need&&(need<3)) need=3;
	if (need>cnt) {puts("No");return 0;}
	if (m<need) {puts("No");return 0;}
	if (m>(ll)cnt*(cnt-1)/2) {puts("No");return 0;}
	puts("Yes");
	return 0;
}