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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;



class UnionFind{
    public:
    //親の番号を格納する。親だった場合は-(その集合のサイズ)
    vector<int> parent;

    //-1で初期化。最初はすべてバラバラ
    UnionFind(int N){
        parent = vector<int>(N,-1);
    }

    //Aがどのグループに属しているか(Aの根)を調べる
    int root(int A){
        if(parent[A] < 0) return A;
        return parent[A]=root(parent[A]);
    }

    //自分のいるグループの頂点数を調べる
    int size(int A){
        return -parent[root(A)];
    }

    //AとBをくっつける
    bool unite(int A, int B) {
    //AとBを直接つなぐのではなく、root(A)にroot(B)をくっつける
        A = root(A);
        B = root(B);
        //既にくっついているからくっつけない
        if(A == B) return false; 

        //大きいほう(A)に小さいほう(B)をくっつける
        //大小が逆ならひっくり返す
        if(size(A) < size(B)) swap(A,B);

        //Aのサイズを更新する
        parent[A] += parent[B];
        //Bの親をAに変更する
        parent[B] = A;

        return true;
    }

    //AとBが同じグループならtrueを返す
    bool same(int A, int B){
        return root(A)==root(B);
    } 
};

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(20);

    ll n,m,q;
    cin>>n>>m>>q;
    UnionFind uni(n);
    vector<pair<int,int>> v;
    for(int i=0;i<q;i++){
        int a,b,c;
        cin>>a>>b>>c;
        if(c==0) uni.unite(a,b);
        else v.push_back(make_pair(a,b));
    }
    ll cnt=0,siz=0;
    for(int i=0;i<n;i++){
        if(i==uni.root(i)){
            cnt++;
            siz += uni.size(i)-1;
        }
    }

    for(auto i:v){
        if(uni.same(i.first,i.second)){
            cout << "No" << endl;
            return 0;
        }
    }
    if(v.size() && cnt<=2){
        cout << "No" << endl;
        return 0;
    }
    if(m>=siz+cnt-1 && m<=siz + cnt*(cnt-1)/2){
        cout << "Yes" << endl;
    }
    else cout << "No" << endl;
    // cerr << siz << " " << cnt << endl;
    // cerr << siz+cnt-1 << " " << m << " " << siz + cnt*(cnt-1)/2 << endl;




}