#include <iostream>
#include <vector>
using namespace std;
using ll = long long;

class DSU {
	private:
		vector<int> comp, siz;
	public:
		DSU(int n) : comp(n), siz(n, 1) {
			for (int i = 0; i < n; ++i) comp[i] = i;
		}
		int get(int i) {
			if (comp[i] != i) comp[i] = get(comp[i]);
			return comp[i];
		}
		bool join(int a, int b) {
			a = get(a);
			b = get(b);
			if (a == b) return false;

			if (siz[a] < siz[b]) swap(a, b);
			siz[a] += siz[b];
			comp[b] = a;
			return true;
		}
		int compSize(int i) { return siz[get(i)]; }
};

bool solve() {
	int n, q;
	ll m;
	cin >> n >> m >> q;

	int cc = n; // component count
	DSU dsu(n);
	vector<pair<int, int>> bans;
	for (int i = 0; i < q; ++i) {
		int a, b, c;
		cin >> a >> b >> c;
		if (c == 0) cc -= dsu.join(a, b);
		else bans.emplace_back(a, b);
	}
	for (auto pr : bans) {
		if (dsu.get(pr.first) == dsu.get(pr.second)) return false;
	}
	ll low = n-cc + cc;
	ll high = n-cc + (ll)cc * (cc - 1) / 2;
	if (cc == 1 || bans.empty()) --low;
	return (low <= m && m <= high);
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	bool res = solve();
	cout << (res ? "Yes" : "No") << '\n';
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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

### Response

```cpp
#include <iostream>
#include <vector>
using namespace std;
using ll = long long;

class DSU {
	private:
		vector<int> comp, siz;
	public:
		DSU(int n) : comp(n), siz(n, 1) {
			for (int i = 0; i < n; ++i) comp[i] = i;
		}
		int get(int i) {
			if (comp[i] != i) comp[i] = get(comp[i]);
			return comp[i];
		}
		bool join(int a, int b) {
			a = get(a);
			b = get(b);
			if (a == b) return false;

			if (siz[a] < siz[b]) swap(a, b);
			siz[a] += siz[b];
			comp[b] = a;
			return true;
		}
		int compSize(int i) { return siz[get(i)]; }
};

bool solve() {
	int n, q;
	ll m;
	cin >> n >> m >> q;

	int cc = n; // component count
	DSU dsu(n);
	vector<pair<int, int>> bans;
	for (int i = 0; i < q; ++i) {
		int a, b, c;
		cin >> a >> b >> c;
		if (c == 0) cc -= dsu.join(a, b);
		else bans.emplace_back(a, b);
	}
	for (auto pr : bans) {
		if (dsu.get(pr.first) == dsu.get(pr.second)) return false;
	}
	ll low = n-cc + cc;
	ll high = n-cc + (ll)cc * (cc - 1) / 2;
	if (cc == 1 || bans.empty()) --low;
	return (low <= m && m <= high);
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	bool res = solve();
	cout << (res ? "Yes" : "No") << '\n';
}

```