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;
using UL = unsigned int;
using ULL = unsigned long long;
using LL = long long;
#define rep(i, n) for(UL i = 0; i < (n); i++)

struct UF {
	UL N; vector<UL> V;
	UL Z;
	void init(UL n) {
		N = n;
		V.resize(N);
		Z = N;
		rep(i, N) V[i] = i;
	}
	UL root(UL a) {
		if (V[a] == a) return a;
		return V[a] = root(V[a]);
	}
	void unite(UL a, UL b) {
		a = root(a), b = root(b);
		if (a != b) Z--;
		V[a] = b;
	}
};

struct Problem {
public:

	LL N, M, Q;
	UF F;
	vector<pair<UL, UL>> Q1;
	vector<pair<UL, UL>> Q2;

	void Solve() {
		scanf("%lld%lld%lld", &N, &M, &Q);
		F.init(N);
		rep(i, Q) {
			UL a, b, c; scanf("%u%u%u", &a, &b, &c);
			if (c == 0) Q1.push_back({ a, b });
			else Q2.push_back({ a, b });
		}
		for (auto& p : Q1) F.unite(p.first, p.second);
		for (auto& e : Q2) if (F.root(e.first) == F.root(e.second)) { printf("No"); return; }
		LL X = F.Z;
		if (M > (N - 1) + (X - 2) * (X - 1) / 2) { printf("No"); return; }
		if (M < (N - 1)) { printf("No"); return; }
		if (Q2.size()) {
			if (M < (N)) { printf("No"); return; }
			if(F.Z <= 2) { printf("No"); return; }
		}
		printf("Yes");
	}
};

int main() {
	unique_ptr<Problem> p(new Problem());
	p->Solve();
	return 0;
}