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;

const int N=1e5+5;
typedef pair<int,int> pii;
int n,q,t,nt;
long long mn,mx,m;
vector<pii> pt[2];
int p1[N],p2[N],gp[N];

int fp1(int x){
	return (p1[x]==x)?x:(p1[x]=fp1(p1[x]));
}

void lnk1(int x,int y){
	x=fp1(x);y=fp1(y);
	p1[x]=y;
}

int fp2(int x){
	return (p2[x]==x)?x:(p2[x]=fp2(p2[x]));
}

void lnk2(int x,int y){
	x=fp2(x);y=fp2(y);
	p2[x]=y;
}

int main(){
	scanf("%d%I64d%d",&n,&m,&q);
	for(int i=1;i<=q;i++){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		a++;b++;
		pt[c].push_back(make_pair(a,b));
	}
	for(int i=1;i<=n;i++) p1[i]=i;
	for(int i=0;i<pt[0].size();i++) lnk1(pt[0][i].first,pt[0][i].second);
	for(int i=1;i<=n;i++) if(fp1(i)==i) t++;
	mn=mx=n-t;
	mx+=1ll*t*(t-1)/2;
	for(int i=1;i<=n;i++) p2[i]=i;
	for(int i=0;i<pt[1].size();i++){
		int x=pt[1][i].first;
		int y=pt[1][i].second;
		if(p1[x]==p1[y]){
			puts("No");
			return 0;
		}
		lnk2(p1[x],p1[y]);
	}
	for(int i=1;i<=n;i++){
		if(fp1(i)==i&&fp2(i)==i) nt++;
		gp[p2[i]]++;
	}
	for(int i=1;i<=n;i++) if(gp[i]>1) mn+=gp[i];
	mn+=nt-1;
//	printf("%lld %lld\n",mn,mx);
	if(mn<=m&&m<=mx) puts("Yes");
	else puts("No");
	return 0;
}