#include <cstdio>

typedef long long LL;
const int MN = 100005, MQ = 100005;

int N, Q; LL M;
int pa[MN]; int fp(int x) { return pa[x] ? pa[x] = fp(pa[x]) : x; }
int qu[MQ], qv[MQ], K;
int E;

int main() {
	scanf("%d%lld%d", &N, &M, &Q);
	for (int i = 1; i <= Q; ++i) {
		int x, y, z;
		scanf("%d%d%d", &x, &y, &z);
		++x, ++y;
		if (z == 0) { if ((x = fp(x)) != (y = fp(y))) pa[y] = x, ++E; }
		else qu[++K] = x, qv[K] = y;
	}
	if (K && M == N - 1) return puts("No"), 0;
	for (int i = 1; i <= K; ++i)
		if ((qu[i] = fp(qu[i])) == (qv[i] = fp(qv[i])))
			return puts("No"), 0;
	int C = N - E;
	if (C == 2) return puts("No"), 0;
	if (M > E + (LL)C * (C - 1) / 2) return puts("No"), 0;
	return puts("Yes"), 0;
	return 0;
}

### 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 <cstdio>

typedef long long LL;
const int MN = 100005, MQ = 100005;

int N, Q; LL M;
int pa[MN]; int fp(int x) { return pa[x] ? pa[x] = fp(pa[x]) : x; }
int qu[MQ], qv[MQ], K;
int E;

int main() {
	scanf("%d%lld%d", &N, &M, &Q);
	for (int i = 1; i <= Q; ++i) {
		int x, y, z;
		scanf("%d%d%d", &x, &y, &z);
		++x, ++y;
		if (z == 0) { if ((x = fp(x)) != (y = fp(y))) pa[y] = x, ++E; }
		else qu[++K] = x, qv[K] = y;
	}
	if (K && M == N - 1) return puts("No"), 0;
	for (int i = 1; i <= K; ++i)
		if ((qu[i] = fp(qu[i])) == (qv[i] = fp(qv[i])))
			return puts("No"), 0;
	int C = N - E;
	if (C == 2) return puts("No"), 0;
	if (M > E + (LL)C * (C - 1) / 2) return puts("No"), 0;
	return puts("Yes"), 0;
	return 0;
}

```