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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define int long long

int p[100005];
int n;
long long m;
int q;

vector<pair<int,int> > edges;

int root(int a){
    return p[a]==a?a:root(p[a]);
}



 main(){
    scanf("%lld%lld%lld",&n,&m,&q);
    for (int x = 0; x<n; x++){
        p[x] = x;
    }
    int components = n;
    for (int x = 0; x<q; x++){
        int a,b,c;
        scanf("%lld%lld%lld",&a,&b,&c);
        if (c==0){
            if (root(a)!=root(b)){
                p[root(a)] = root(b);
                components--;
            }

        }
        else{
            edges.push_back({a,b});
        }
    }
    for (auto x : edges){
        if (root(x.first)==root(x.second)){
            printf("No");
            return 0;
        }
    }
    int mn = n-1;
    if (!edges.empty()) mn = n;

    long long mx = components*(components-1)/2 + n-components;

    if (m<mn || m>mx){
        printf("No");
    }
    else{
        printf("Yes");
    }


}