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;

const int N = 1e5 + 5;

int a[N], b[N], c[N], p[N];

int find(int x) {
  return p[x] < 0 ? x : p[x] = find(p[x]);
}

bool merge(int u, int v) {
  if ((u = find(u)) == (v = find(v)))
    return false;
  if (-p[u] > -p[v])
    swap(u, v);
  p[v] += p[u];
  p[u] = v;
  return true;
}

int main() {
  memset(p, -1, sizeof p);
  int n, m, q;
  scanf("%d %d %d", &n, &m, &q);
  int cnt = n;
  int tot = 0;
  for (int i = 0; i < q; ++i) {
    scanf("%d %d %d", a+i, b+i, c+i);
    if (c[i] == 0) {
      tot += merge(a[i], b[i]);
    }
  }
  cnt -= tot;
  set<int> st;
  for (int i = 0;i < q; ++i) {
    if (c[i]) {
      if (find(a[i]) == find(b[i])) {
        puts("No");
        return 0;
      }
      st.insert(find(a[i]));
      st.insert(find(b[i]));
    }
  }
  long long mini = tot + (int)st.size();
  long long maxi = tot + 1LL * cnt * (cnt - 1) / 2;
  puts((mini <= m && m <= maxi) ? "Yes" : "No");
  return 0;
}