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 <cstdio>

#include <algorithm>
#include <functional>
#include <vector>

using namespace std;

typedef long long ll;

int main() {
  int n, q;
  ll m;
  scanf("%d%lld%d", &n, &m, &q);
  vector<int> a(q), b(q), c(q), f(n);
  for (int i = 0; i < n; ++i)
    f[i] = i;
  for (int i = 0; i < q; ++i)
    scanf("%d%d%d", &a[i], &b[i], &c[i]);
  function<int(int)> find = [&](int x) {
    return f[x] == x ? x : f[x] = find(f[x]);
  };
  int cnt = n;
  for (int i = 0; i < q; ++i) {
    if (c[i] == 0) {
      int x = find(a[i]), y = find(b[i]);
      if (x != y) {
        if (x > y) swap(x, y);
        f[y] = x;
        --cnt;
      }
    }
  }
  if (cnt == 2) {
    for (int i = 1; i < n; ++i)
      if (find(i) != 0) {
        f[i] = 0;
        break;
      }
    cnt = 1;
  }
  if (n - cnt + cnt * (cnt - 1LL) / 2 < m) {
    puts("No");
    return 0;
  }
  if (m == n - 1) {
    puts(count(c.begin(), c.end(), 1) ? "No" : "Yes");
    return 0;
  }
  for (int i = 0; i < q; ++i)
    if (c[i] && find(a[i]) == find(b[i])) {
      puts("No");
      return 0;
    }
  puts("Yes");
  return 0;
}