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 popCnt(x) (__builtin_popcountll(x))
typedef long long Long;

const int N = 1e5 + 5;

vector<int> adj[2][N];

int comp[N];

void dfs(int node, int id) {
  if (comp[node] == id) return;
  comp[node] = id;
  for (int x : adj[0][node]) {
    dfs(x, id);
  }
}

int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#ifdef Local
  freopen("test.in", "r", stdin);
#else
#define endl '\n'
#endif

  int n, q;
  Long m;
  cin >> n >> m >> q;

  bool type_1_appeared = false;
  while (q--) {
    int u, v, c;
    cin >> u >> v >> c;
    ++u, ++v;
    adj[c][u].push_back(v);
    adj[c][v].push_back(u);
    type_1_appeared |= (c == 1);
  }

  int cnt_comps = 0;
  for (int i = 1; i <= n; ++i) {
    if (comp[i] == 0) {
      dfs(i, i);
      ++cnt_comps;
    }
  }

  for (int i = 1; i <= n; ++i) {
    for (int x : adj[1][i]) {
      if (comp[i] == comp[x]) {
        cout << "No" << endl;
        return 0;
      }
    }
  }

  int mn = n;
  Long mx = n - cnt_comps + 1LL * cnt_comps * (cnt_comps - 1) / 2;
  if (!type_1_appeared) {
    mn = n - 1;
  }
  if (cnt_comps == 2 && type_1_appeared) {
    cout << "No" << endl;
    return 0;
  }

  if (mn <= m && m <= mx) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }

}