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 lint;

const int MAXN = 100010;

int U[MAXN], V[MAXN], C[MAXN];
int uni[MAXN], sz[MAXN];

int guni(int x) { return x == uni[x] ? x : uni[x] = guni(uni[x]); }
void unite(int x, int y) { uni[guni(x)] = guni(y); }

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	int N, Q;
	lint M;

	cin >> N >> M >> Q;
	for(int i = 0; i < Q; i++) cin >> U[i] >> V[i] >> C[i];

	for(int i = 0; i < N; i++) uni[i] = i;
	for(int i = 0; i < Q; i++) if(C[i] == 0) unite(U[i], V[i]);
	for(int i = 0; i < Q; i++) if(C[i] == 1 && guni(U[i]) == guni(V[i])) {
		cout << "No";
		return 0;
	}
	int cnt = 0;
	for(int i = 0; i < N; i++) if(i == uni[i]) cnt++;
	if(M > lint(cnt) * (cnt - 1) / 2 + N - cnt) {
		cout << "No";
		return 0;
	}
	for(int i = 0; i < Q; i++) if(C[i] == 1 && (M == N - 1 || cnt == 2)) {
		cout << "No";
		return 0;
	}
	if(cnt == 2 && M != N - 1) {
		cout << "No";
		return 0;
	}
	cout << "Yes";
	return 0;
}