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:
// IOI 2021
#include <bits/stdc++.h>
using namespace std;

#define endl '\n'
#define ends ' '
#define die(x) return cout << x << endl, 0
#define all(v) v.begin(), v.end()
#define sz(x) (int)(x.size())
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) { cerr << ends << H; debug_out(T...); }
#define debug(...) cerr << "{" << #__VA_ARGS__ << "}:", debug_out(__VA_ARGS__)
typedef long long ll;
typedef pair<int, int> pii;
const int INF = 1e9;
const ll MOD = 998244353;
 
////////////////////////////////////////////////////////////////////

const int N = 1e5 + 5;

int n, q, c, P[N];
ll m;
vector<pii> E;

int GetPar(int v) { return P[v] == v ? v : P[v] = GetPar(P[v]); }
void Merge(int v, int u) {
	v = GetPar(v), u = GetPar(u);
	if (v == u) return;
	P[v] = u;
	c--;
}

int main() {

	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	mt19937 Rnd(time(0));
	
	cin >> n >> m >> q; c = n;
	for (int i = 0; i < n; i++) P[i] = i;
	while (q--) {
		int v, u, c; cin >> v >> u >> c;
		if (c) E.push_back(make_pair(v, u));
		else Merge(v, u);
	}
	if (m == n - 1) die((sz(E) ? "No" : "Yes"));
	for (pii e : E) if (GetPar(e.first) == GetPar(e.second)) die("No");
	cout << (m > n - c + 1LL * c * (c - 1) / 2 ? "No" : "Yes") << endl;

	return 0;
}