#include <iostream>
#include <set>
#include <vector>
#define llint long long

using namespace std;

struct UnionFind{
	int size;
	vector<int> parent;
	
	UnionFind(){}
	UnionFind(int size){
		this->size = size;
		parent.resize(size+1);
		init();
	}
	void init(){
		for(int i = 0; i <= size; i++) parent[i] = i;
	}
	int root(int i){
		if(parent[i] == i) return i;
		return parent[i] = root(parent[i]);
	}
	bool same(int i, int j){
		return root(i) == root(j);
	}
	void unite(int i, int j){
		int root_i = root(i), root_j = root(j);
		if(root_i == root_j) return;
		parent[root_i] = root_j;
	}
};

llint n, m, Q;
llint u[100005], v[100005], t[100005];
set<llint> S;
UnionFind uf(100005);

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n >> m >> Q;
	for(int i = 1; i <= Q; i++) cin >> u[i] >> v[i] >> t[i];
	
	llint cnt = 0;
	for(int i = 1; i <= Q; i++){
		if(t[i] == 0) uf.unite(u[i], v[i]), cnt++;
	}
	for(int i = 1; i <= Q; i++){
		if(t[i] == 1 && uf.same(u[i], v[i])){
			cout << "No" << endl;
			return 0;
		}
	}
	
	for(int i = 0; i < n; i++) S.insert(uf.root(i));
	llint c = S.size();
	
	llint l = n - 1, r = n-c + c*(c-1)/2;
	
	if(cnt == Q){
		if(l <= m && m <= r) cout << "Yes" << endl;
		else cout << "No" << endl;
	}
	else{
		if(l < m && m <= r) cout << "Yes" << endl;
		else cout << "No" << endl;
	}
}
### Prompt

Generate a cpp solution to the following 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

### Response

```cpp
#include <iostream>
#include <set>
#include <vector>
#define llint long long

using namespace std;

struct UnionFind{
	int size;
	vector<int> parent;
	
	UnionFind(){}
	UnionFind(int size){
		this->size = size;
		parent.resize(size+1);
		init();
	}
	void init(){
		for(int i = 0; i <= size; i++) parent[i] = i;
	}
	int root(int i){
		if(parent[i] == i) return i;
		return parent[i] = root(parent[i]);
	}
	bool same(int i, int j){
		return root(i) == root(j);
	}
	void unite(int i, int j){
		int root_i = root(i), root_j = root(j);
		if(root_i == root_j) return;
		parent[root_i] = root_j;
	}
};

llint n, m, Q;
llint u[100005], v[100005], t[100005];
set<llint> S;
UnionFind uf(100005);

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n >> m >> Q;
	for(int i = 1; i <= Q; i++) cin >> u[i] >> v[i] >> t[i];
	
	llint cnt = 0;
	for(int i = 1; i <= Q; i++){
		if(t[i] == 0) uf.unite(u[i], v[i]), cnt++;
	}
	for(int i = 1; i <= Q; i++){
		if(t[i] == 1 && uf.same(u[i], v[i])){
			cout << "No" << endl;
			return 0;
		}
	}
	
	for(int i = 0; i < n; i++) S.insert(uf.root(i));
	llint c = S.size();
	
	llint l = n - 1, r = n-c + c*(c-1)/2;
	
	if(cnt == Q){
		if(l <= m && m <= r) cout << "Yes" << endl;
		else cout << "No" << endl;
	}
	else{
		if(l < m && m <= r) cout << "Yes" << endl;
		else cout << "No" << endl;
	}
}
```