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 <iostream>
#include <vector>

using namespace std;

void success () {
  cout << "Yes" << endl;
  exit(0);
}

void failure () {
  cout << "No" << endl;
  exit(0);
}

const int MAX_N = 1e5 + 5;

int root [MAX_N];

typedef long long ll;

ll sq (ll x) {
  return x * (x - 1) / 2;
}

int find (int u) {
  if (root[u] != u) {
    root[u] = find(root[u]);
  }
  return root[u];
}

void merge (int u, int v) {
  u = find(u);
  v = find(v);

  root[u] = v;
}

int main () {
  ios::sync_with_stdio(false);

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

  for (int i = 0; i < n; i++) {
    root[i] = i;
  }

  vector<pair<int, int>> ones;
  for (int i = 0; i < q; i++) {
    int u, v, c;
    cin >> u >> v >> c;

    if (c == 0) {
      merge(u, v);
    } else {
      ones.push_back(make_pair(u, v));
    }
  }

  int compc = 0;
  for (int i = 0; i < n; i++) {
    if (root[i] == i) {
      compc++;
    }
  }

  if (compc <= 2) {
    if (m != n - 1 || !ones.empty()) {
      failure();
    } else {
      success();
    }
  }

  for (auto pr : ones) {
    int u = pr.first;
    int v = pr.second;

    if (find(u) == find(v)) {
      failure();
    }
  }

  if (m == n - 1 && !ones.empty()) {
    failure();
  }

  ll maxe = (n - compc) + sq(compc);
  if (m > maxe) {
    failure();
  }

  success();
}