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;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);

  long N,M,Q; cin>>N>>M>>Q;

  vector<int> par(N);
  iota(par.begin(), par.end(), 0);
  auto root = [f=[&](auto g, int x) {
    if (par[x] == x)
      return x;
    return par[x] = g(g, par[x]);
  }](int x) {
    return f(f, x);
  };
  auto unite = [&](int x, int y) {
    return root(y) != (par[root(y)] = root(x));
  };

  vector<pair<int, int>> sconds, dconds;
  for (int q = 0; q < Q; q++) {
    int a, b, c; cin>>a>>b>>c;
    if (c == 0)
      sconds.emplace_back(a, b);
    else
      dconds.emplace_back(a, b);
  }
  int c_graph = N;
  for (auto cond : sconds) {
    if (unite(cond.first, cond.second))
      c_graph--;
  }
  for (auto cond : dconds) {
    if (root(cond.first) == root(cond.second)) {
      cout << "No" << endl;
      return 0;
    }
  }
  int c_min = dconds.empty() ? N-1 : N;
  bool ans = (M >= c_min) and (M <= (long)N + c_graph*(c_graph-3)/2);
  cout << (ans ? "Yes" : "No") << endl;

  return 0;
}