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:
/**
 *    author:  tourist
 *    created: 21.09.2019 16:14:18       
**/
#include <bits/stdc++.h>

using namespace std;

class dsu {
 public:
  vector<int> p;
  int n;

  dsu(int _n) : n(_n) {
    p.resize(n);
    iota(p.begin(), p.end(), 0);
  }

  inline int get(int x) {
    return (x == p[x] ? x : (p[x] = get(p[x])));
  }

  inline bool unite(int x, int y) {
    x = get(x);
    y = get(y);
    if (x != y) {
      p[x] = y;
      return true;
    }
    return false;
  }
};

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  long long m;
  cin >> n >> m >> q;
  vector<int> x(q), y(q), z(q);
  for (int i = 0; i < q; i++) {
    cin >> x[i] >> y[i] >> z[i];
  }
  dsu d(n);
  int cnt1 = 0;
  int ncomp = n;
  for (int i = 0; i < q; i++) {
    if (z[i] == 0) {
      if (d.unite(x[i], y[i])) {
        ncomp--;
      }
    } else {
      ++cnt1;
    }
  }
  for (int i = 0; i < q; i++) {
    if (z[i] == 1 && d.get(x[i]) == d.get(y[i])) {
      cout << "No" << '\n';
      return 0;
    }
  }
  long long L = n;
  long long R = (long long) ncomp * (ncomp - 1) / 2 + (n - ncomp);
  if (cnt1 == 0) {
    --L;
  }
  if (L <= m && m <= R) {
    cout << "Yes" << '\n';
  } else {
    cout << "No" << '\n';
  }
  return 0;
}