// sync

#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>

const int N = 1e5 + 51;

int p[N];
int f(int x) { return x[p] ? x[p] = f(x[p]) : x; }

int n, m, q, op[N], x[N], y[N], cnt, c0;
int main() {
	scanf("%d%d%d", &n, &m, &q), cnt = n;
	for (int i = 1; i <= q; i++) {
		scanf("%d%d%d", x + i, y + i, op + i), x[i]++, y[i]++;
		int u = x[i], v = y[i];
		if (!op[i]) {
			c0++;
			if ((u = f(u)) != (v = f(v))) u[p] = v, cnt--;
		}
	}
	for (int i = 1; i <= q; i++)
		if (op[i] && f(x[i]) == f(y[i])) return puts("No"), 0;
	if (n - (c0 == q) <= m && m <= n - cnt + 1ll * cnt * (cnt - 1) / 2)
		puts("Yes");
	else
		puts("No");
}
### Prompt

Construct a CPP code solution to the problem outlined:

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
// sync

#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>

const int N = 1e5 + 51;

int p[N];
int f(int x) { return x[p] ? x[p] = f(x[p]) : x; }

int n, m, q, op[N], x[N], y[N], cnt, c0;
int main() {
	scanf("%d%d%d", &n, &m, &q), cnt = n;
	for (int i = 1; i <= q; i++) {
		scanf("%d%d%d", x + i, y + i, op + i), x[i]++, y[i]++;
		int u = x[i], v = y[i];
		if (!op[i]) {
			c0++;
			if ((u = f(u)) != (v = f(v))) u[p] = v, cnt--;
		}
	}
	for (int i = 1; i <= q; i++)
		if (op[i] && f(x[i]) == f(y[i])) return puts("No"), 0;
	if (n - (c0 == q) <= m && m <= n - cnt + 1ll * cnt * (cnt - 1) / 2)
		puts("Yes");
	else
		puts("No");
}
```