#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7;

int u[N], v[N], w[N], f[N];
bool vis[N];

int find(int x){
	if (f[x] == x) return x;
	return f[x] = find(f[x]);
}

int main(){
	int n, q; long long m;
	scanf("%d%lld%d", &n, &m, &q);
	for (int i = 1; i <= n; i++) f[i] = i;
	int n1 = n;
	for (int i = 1; i <= q; i++){
		scanf("%d%d%d", &u[i], &v[i], &w[i]); u[i] ++; v[i] ++;
		if (w[i] == 0){
			if (find(u[i]) == find(v[i])) continue;
			n1 --;
			f[find(u[i])] = find(v[i]);
		}
	}

	int num = 0;
	for (int i = 1; i <= q; i++){
		if (w[i] == 1){
			if (find(u[i]) == find(v[i])) return 0 * printf("No");
			num = 1;
		}
	}
	if (num > 0 && n1 < 3) return 0 * printf("No");
	long long m1 = n - 1;
	if (num > 0) m1 ++;
	long long m2 = 1LL * n1 * (n1 - 1) / 2 + n - n1;
	if (m >= m1 && m <= m2) return 0 * printf("Yes");
	else return 0 * printf("No");
	return 0;
}
### Prompt

Create a solution in Cpp for 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 <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7;

int u[N], v[N], w[N], f[N];
bool vis[N];

int find(int x){
	if (f[x] == x) return x;
	return f[x] = find(f[x]);
}

int main(){
	int n, q; long long m;
	scanf("%d%lld%d", &n, &m, &q);
	for (int i = 1; i <= n; i++) f[i] = i;
	int n1 = n;
	for (int i = 1; i <= q; i++){
		scanf("%d%d%d", &u[i], &v[i], &w[i]); u[i] ++; v[i] ++;
		if (w[i] == 0){
			if (find(u[i]) == find(v[i])) continue;
			n1 --;
			f[find(u[i])] = find(v[i]);
		}
	}

	int num = 0;
	for (int i = 1; i <= q; i++){
		if (w[i] == 1){
			if (find(u[i]) == find(v[i])) return 0 * printf("No");
			num = 1;
		}
	}
	if (num > 0 && n1 < 3) return 0 * printf("No");
	long long m1 = n - 1;
	if (num > 0) m1 ++;
	long long m2 = 1LL * n1 * (n1 - 1) / 2 + n - n1;
	if (m >= m1 && m <= m2) return 0 * printf("Yes");
	else return 0 * printf("No");
	return 0;
}
```