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;

#define range(i, m, n) for(int i = m; i < n; i++)
#define husk(i, m, n) for(int i = m; i > n; i--)

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 n, q;
long long m;

int main() {
  cin >> n >> m >> q;
  vector<pair<int, int>> s[2];
  while(q--) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    s[c].emplace_back(a, b);
  }
  dsu ss(n);
  int comp = n;
  for(auto u : s[0]) {
    comp -= ss.unite(u.first, u.second);
  }
  for(auto u : s[1]) {
    if(ss.get(u.first) == ss.get(u.second)) {
      cout << "No";
      return 0;
    }
  }
  long long L = n - (s[1].empty());
  long long R = (long long) comp * (comp - 1) / 2 + (n - comp);
  if(L <= m && m <= R) cout << "Yes";
  else cout << "No";
  return 0;
}