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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.*;

public class Main{
    public static void main(String[] args){
    //入力
        Scanner sc = new Scanner(System.in);
        int N =sc.nextInt();
        int M =sc.nextInt();
        int Q =sc.nextInt();
        int[] solo = new int[N];
        for(int i=0;i<N;i++){
            solo[i] = -1;
        }
        for(int i=0;i<Q;i++){
            int t = sc.nextInt();
            int tt = sc.nextInt();
            int ttt = sc.nextInt();
            if(ttt==0){
                if(solo[t]==-1){solo[t] = 0;}
                if(solo[tt]==-1){solo[tt] = 0;}
            }
            else{
                solo[t] = 1;
                solo[tt] = 1;
            }
        }
    //処理
        int a=0;
        int b=0;
        int c=0;
        boolean ans = false;
        for(int i=0;i<N;i++){
            if(solo[i]==1){a++;}
            if(solo[i]==0){b++;}
            if(solo[i]==-1){c++;}
        }
        int E = M - b;
        N = a + c;
        if(3 <= a+c){
            for(int i=0;i<=b;i++){
                if((long)(E-i)<=((long)(N-i)*(long)(N-i-1)/2L)){
                    if(N-i != 2){ans = true;}
                }
            }
        }
        System.out.println((ans)?"Yes":"No");
    }
}