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<iostream>
#include<cstring>
#include<cassert>
#include<cmath>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<time.h>
#include<algorithm>
using namespace std;
#define REP(i,x,y) for(ll i=x;i<=y;i++)
#define rep(i,n) REP(i,1,n)
#define rep0(i,n) REP(i,0,n-1)
#define repG(i,x) for(ll i=pos[x];~i;i=e[i].next)
#define ll long long
#define db double
const ll N=1e5+7;
const ll INF=1e9+7;
ll n,m,q;
ll a[N],b[N],c[N],f[N],sz[N];
ll ff(ll x){return (f[x]==x)?x:f[x]=ff(f[x]);}

int main(){
	scanf("%lld%lld%lld",&n,&m,&q);
	rep(i,q)scanf("%lld%lld%lld",&a[i],&b[i],&c[i]);
	rep(i,q)a[i]++,b[i]++;
	if(m==n-1)rep(i,q)if(c[i]){puts("No"); return 0;}
	rep(i,n)f[i]=i,sz[i]=1;
	rep(i,q){
		if(!c[i]){
			ll u=ff(a[i]),v=ff(b[i]);
			if(u!=v){
				sz[v]+=sz[u];
				f[u]=v;
			}
		}
	}
	rep(i,q)if(c[i])if(ff(a[i])==ff(b[i])){puts("No"); return 0;}
	ll hh=0;
	rep(i,n)if(ff(i)==i)hh++;
	if(m>n+hh*(hh-3)/2){puts("No"); return 0;}
	puts("Yes");
	return 0;
}