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;

typedef long long ll;
ll N, M;
ll Q;
vector<pair<ll, ll>> Black;

const ll MAX_N = 100000;
ll Parent[MAX_N], Rank[MAX_N], NGroups;
void init() {
  fill(Parent, Parent + N, -1);
  fill(Rank, Rank + N, 1);
  NGroups = N;
}
ll root(ll x) {
  if (Parent[x] == -1) return x;
  return Parent[x] = root(Parent[x]);
}
void unite(ll x, ll y) {
  x = root(x);
  y = root(y);
  if (x == y) return;
  --NGroups;
  if (Rank[x] > Rank[y]) {
    Parent[y] = x;
  } else {
    Parent[x] = y;
    if (Rank[x] == Rank[y]) {
      ++Rank[y];
    }
  }
}
bool same(ll x, ll y) { return root(x) == root(y); }

bool solve() {
  for (ll i = 0; i < Q; ++i) {
    ll a, b, c;
    cin >> a >> b >> c;
    if (c == 0) {
      unite(a, b);
    } else {
      Black.push_back(make_pair(a, b));
    }
  }
  for (auto pair : Black) {
    if (M == N - 1) return false;
    if (same(pair.first, pair.second)) return false;
  }
  return M - N <= NGroups * (NGroups - 3) / 2;
}
int main() {
  cin >> N >> M >> Q;
  init();
  cout << (solve() ? "Yes" : "No") << endl;
  return 0;
}