#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>

using namespace std;

typedef long long ll;

const int maxn = 1e5 + 10;

int root[maxn];

int get_root(int x){
	return root[x] = root[x] == x ? x : get_root(root[x]);
}

int N , Q;
ll M;
int l[maxn] , r[maxn] , cnt = 0;
int v[maxn];

void output(int x){
	cout << (x ? "Yes" : "No") << endl;
	exit(0);
}

int main(){
	cin >> N >> M >> Q;
	for(int i = 0 ; i < N ; ++ i)
		root[i] = i;
	for(int i = 1 ; i <= Q ; ++ i){
		int a , b , c;
		cin >> a >> b >> c;
		if(c == 0){
			int a1 = get_root(a) , b1 = get_root(b);
			if(a1 != b1)
				root[a1] = b1;
		}
		else{
			++ cnt;
			l[cnt] = a , r[cnt] = b;
		}
	}
	int sum = 0;
	for(int i = 0 ; i < N ; ++ i)
		if(root[i] == i)
			sum ++;
	if(sum == 1){
		if(cnt > 0)
			output(0);
		else if(M != N - 1)
			output(0);
		output(1);
	}
	else if(sum == 2){
		if(cnt > 0)
			output(0);
		else if(M != N - 2)
			output(0);
		output(1);
	}
	else{
		for(int i = 1 ; i <= cnt ; ++ i){
			int a1 = get_root(l[i]) , b1 = get_root(r[i]);
			if(a1 == b1)
				output(0);
			v[a1] = v[b1] = 1;
		}
		int s1 = 0;
		for(int i = 0 ; i < N ; ++ i)
			if(v[i] > 0)
				s1 ++;
		ll MINN = N - sum + (s1 == 2 ? 3 : s1) , MAXN = N - sum + 1ll * sum * (sum - 1) / 2;
		if(M >= MINN && M <= MAXN)
			output(1);
		output(0); 
	}
	return 0;
}
### Prompt

Please formulate 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 <cstdio>
#include <cstring>
#include <cstdlib>

using namespace std;

typedef long long ll;

const int maxn = 1e5 + 10;

int root[maxn];

int get_root(int x){
	return root[x] = root[x] == x ? x : get_root(root[x]);
}

int N , Q;
ll M;
int l[maxn] , r[maxn] , cnt = 0;
int v[maxn];

void output(int x){
	cout << (x ? "Yes" : "No") << endl;
	exit(0);
}

int main(){
	cin >> N >> M >> Q;
	for(int i = 0 ; i < N ; ++ i)
		root[i] = i;
	for(int i = 1 ; i <= Q ; ++ i){
		int a , b , c;
		cin >> a >> b >> c;
		if(c == 0){
			int a1 = get_root(a) , b1 = get_root(b);
			if(a1 != b1)
				root[a1] = b1;
		}
		else{
			++ cnt;
			l[cnt] = a , r[cnt] = b;
		}
	}
	int sum = 0;
	for(int i = 0 ; i < N ; ++ i)
		if(root[i] == i)
			sum ++;
	if(sum == 1){
		if(cnt > 0)
			output(0);
		else if(M != N - 1)
			output(0);
		output(1);
	}
	else if(sum == 2){
		if(cnt > 0)
			output(0);
		else if(M != N - 2)
			output(0);
		output(1);
	}
	else{
		for(int i = 1 ; i <= cnt ; ++ i){
			int a1 = get_root(l[i]) , b1 = get_root(r[i]);
			if(a1 == b1)
				output(0);
			v[a1] = v[b1] = 1;
		}
		int s1 = 0;
		for(int i = 0 ; i < N ; ++ i)
			if(v[i] > 0)
				s1 ++;
		ll MINN = N - sum + (s1 == 2 ? 3 : s1) , MAXN = N - sum + 1ll * sum * (sum - 1) / 2;
		if(M >= MINN && M <= MAXN)
			output(1);
		output(0); 
	}
	return 0;
}
```