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;
typedef double D;
typedef pair<ll,ll> P;
typedef tuple<ll,ll,ll> T;
#define M 998244353
#define F first
#define S second
#define PB push_back
#define INF 1000000000
int q,par[100005];
ll n,k,m;
vector<P>u,s;
int find(int x){
	if(x==par[x])return x;
	return par[x]=find(par[x]);
}
bool same(int x,int y){
	x=find(x);
	y=find(y);
	return x==y;
}
void unit(int x,int y){
	x=find(x);
	y=find(y);
	par[x]=y;
}
int main(void){
	scanf("%lld%lld%d",&n,&m,&q);
	k=n;
	for(int i=0;i<n;i++)par[i]=i;
	for(int i=0;i<q;i++){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		if(c==0)u.PB(P(a,b));
		else s.PB(P(a,b));
	}
	for(int i=0;i<u.size();i++){
		int a=u[i].F,b=u[i].S;
		if(!same(a,b)){
			k--;
			unit(a,b);
		}
	}
	if(s.size()>0&&m<n){
		printf("No\n");
		return 0;
	}
	for(int i=0;i<s.size();i++){
		int a=s[i].F,b=s[i].S;
		if(same(a,b)){
			printf("No\n");
			return 0;
		}
	}
	if(k*(k-1LL)/2LL+n-k>=m)printf("Yes\n");
	else printf("No\n");
}