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;

long long N, M, Q, A[1 << 17], B[1 << 17], C[1 << 17], cnt, col[1 << 17];
bool flag = false;
vector<int> X[100009];

void dfs(int pos) {
	if (col[pos] >= 1) return;
	col[pos] = cnt;
	for (int i = 0; i < X[pos].size(); i++) dfs(X[pos][i]);
}

int main() {
	cin >> N >> M >> Q;
	for (int i = 0; i < Q; i++) {
		cin >> A[i] >> B[i] >> C[i];
		if (C[i] == 0) {
			X[A[i]].push_back(B[i]);
			X[B[i]].push_back(A[i]);
		}
		if (C[i] == 1) flag = true;
	}
	for (int i = 0; i < N; i++) {
		if (col[i] >= 1) continue;
		cnt++; dfs(i);
	}

	for (int i = 0; i < Q; i++) {
		if (col[A[i]] == col[B[i]] && C[i] == 1) {
			cout << "No" << endl;
			return 0;
		}
	}

	if (flag == true && cnt <= 2) {
		cout << "No" << endl;
	}
	else {
		long long cl = N, cr = (N - cnt) + cnt * (cnt - 1) / 2LL;
		if (flag == false) cl = N - 1;
		if (cl <= M && M <= cr) cout << "Yes" << endl;
		else cout << "No" << endl;
	}
	return 0;
}