Problem: 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`.

Examples

Input

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


Output

Yes


Input

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


Output

No


Input

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


Output

No

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

int main()
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);

  int n, q;
  long long m;
  cin >> n >> m >> q;

  vector<int> ds(n);
  for (int i = 0; i < n; i++)
    ds[i] = i;

  function<int(int)> get = [&](int x)
  {
    return x == ds[x] ? x : ds[x] = get(ds[x]);
  };

  vector<pair<int, int>> queries;
  int dsCnt = n;
  while (q--)
  {
    int x, y, c;
    cin >> x >> y >> c;
    if (c) queries.push_back({x, y});
    else
    {
      int dx = get(x), dy = get(y);
      if (dx != dy)
      {
        ds[dx] = dy;
        dsCnt--;
      }
    }
  }

  if (m == n - 1) cout << (queries.empty() ? "Yes\n" : "No\n");
  else
  {
    for (auto u : queries)
      if (get(u.first) == get(u.second))
      {
        cout << "No\n";
        return 0;
      }

    long long maxM = (n - dsCnt) + (dsCnt - 1LL) * dsCnt / 2;
    cout << (m <= maxM ? "Yes\n" : "No\n");
  }
}