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;

struct dsu {
  int cc;
  vector<int> p, sz;
  dsu(int n) : cc(n), p(n, -1), sz(n, 1) {}
  int root(int v) const {
    while (p[v] != -1) v = p[v];
    return v;
  }
  bool unite(int u, int v) {
    u = root(u), v = root(v);
    if (u == v) return false;
    --cc;
    if (sz[u] < sz[v]) swap(u, v);
    p[v] = u;
    sz[u] += sz[v];
    return true;
  }
  bool same(int u, int v) const { return root(u) == root(v); }
  int size(int v) const { return sz[root(v)]; }
};

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int n, q;
  long long m;
  cin >> n >> m >> q;
  vector<int> a(q), b(q), c(q);
  dsu d(n);
  for (int i = 0; i < q; ++i) {
    cin >> a[i] >> b[i] >> c[i];
    if (c[i] == 0) {
      d.unite(a[i], b[i]);
    }
  }
  for (int i = 0; i < q; ++i) {
    if (c[i] == 0) {
      continue;
    }
    if (d.same(a[i], b[i]) or d.cc < 3 or m == n - 1) {
      cout << "No\n";
      exit(0);
    }
  }
  if (m > (long long)d.cc * (d.cc - 1) / 2 + (n - d.cc)) {
    cout << "No\n";
  } else {
    cout << "Yes\n";
  }
}