Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

5 5 3
0 1 0
1 2 1
2 3 0

Sample Output 1

Yes

For example, consider a graph with edges (0,1),(1,2),(1,4),(2,3),(2,4). This graph matches the clues.

Sample Input 2

4 4 3
0 1 0
1 2 1
2 3 0

Sample Output 2

No

Sample Input 3

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

Sample Output 3

No

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define INF ((1LL<<62)-(1LL<<31))
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
#define bit(n,m)(((n)>>(m))&1)
int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}
ll pom(ll a,ll n,int m){ll x=1;for(a%=m;n;n/=2)n&1?x=x*a%m:0,a=a*a%m;return x;}
//#define MOD 998244353
#define MOD 1000000007
#define invp(a,p)pom(a,p-2,p)

//union-find
#define UFLIMIT (1<<17)
ll unicnt[UFLIMIT+10];//正ならcnt、非正なら根のindex
void ufinit(int n){rep(i,0,n)unicnt[i]=1;}
int ufroot(int x){return unicnt[x]<=0?-(unicnt[x]=-ufroot(-unicnt[x])):x;}
int ufsame(int x,int y){return ufroot(x)==ufroot(y);}
void uni(int x,int y){if((x=ufroot(x))==(y=ufroot(y)))return;if(unicnt[x]<unicnt[y]){int t=x;x=y;y=t;}unicnt[x]+=unicnt[y];unicnt[y]=-x;}
#undef UFLIMIT

int a[100010],b[100010],c[100010];
int main(){
	int n,m,q;
	scanf("%d%d%d",&n,&m,&q);
	rep(i,0,q)scanf("%d%d%d",a+i,b+i,c+i);

	ufinit(n);
	rep(i,0,q)if(c[i])uni(a[i],b[i]);

	rep(i,0,q)if(!c[i]){
		if(!ufsame(a[i],b[i])){
			uni(a[i],b[i]);
			unicnt[ufroot(a[i])]--;
		}
	}

	ll c=0,cnt0=0;
	rep(i,0,n){
		if(unicnt[i]==1)cnt0++;
		else if(unicnt[i]>=2)c+=unicnt[i];
	}
	if(c==2&&cnt0==0)puts("No");
	else puts(!!c+n-1<=m&&m<=(c+cnt0)*(c+cnt0-1)/2+(n-c-cnt0)?"Yes":"No");
}