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>
#include <assert.h>
using namespace std;
using i64 = long long;

struct UnionFind
{
  vector<i64> data;
  i64 cnt;
  UnionFind(i64 size)
  {
    data.resize(size, -1);
    cnt = size;
  }
  bool unite(i64 x, i64 y)
  {
    x = root(x);
    y = root(y);
    if (x != y)
    {
      if (data[y] < data[x])
        swap(x, y);
      data[x] += data[y];
      data[y] = x;
      cnt--;
    }
    return x != y;
  }
  bool same(i64 x, i64 y)
  {
    return root(x) == root(y);
  }
  i64 root(i64 x)
  {
    return (data[x] < 0 ? x : data[x] = root(data[x]));
  }
  i64 size(i64 x)
  {
    return -data[root(x)];
  }
};

int main()
{
  i64 n, m, q;
  cin >> n >> m >> q;
  vector<pair<i64, i64>> d;
  UnionFind uf(n);
  for (i64 _ = 0; _ < q; _++)
  {
    i64 a, b, c;
    cin >> a >> b >> c;
    if (c == 0)
      uf.unite(a, b);
    else
      d.push_back({a, b});
  }
  if (m == n - 1)
  {
    if (d.size() == 0)
      cout << "Yes" << endl;
    else
      cout << "No" << endl;
    return 0;
  }
  for (pair<i64, i64> i : d)
    if (uf.same(i.first, i.second))
    {
      cout << "No" << endl;
      return 0;
    }
  if (m <= n - uf.cnt + uf.cnt * (uf.cnt - 1) / 2)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  return 0;
}