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

Solution (in CPP):

Solution:
#include <cstdio>
#include <vector>
using namespace std;
const int NC = 1e5 + 5;
typedef long long ll;
int N, M, Q, fa[NC];
void Init() {
	for (int i = 1; i <= N; ++i) fa[i] = i;
}
int Get(int u) { return (u == fa[u]) ? u : (fa[u] = Get(fa[u])); }
void Merge(int u, int v) {
	u = Get(u);
	v = Get(v);
	if (u != v)
		fa[u] = v;
}
bool same(int u, int v) {
	if (Get(u) == Get(v))
		return 1;
	else return 0;
}
vector<pair<int, int> > ns;
int main() {
	scanf("%d%d%d", &N, &M, &Q);
	Init();
	bool hav1 = 0;
	for (int i = 1; i <= Q; ++i) {
		int A, B, C;
		scanf("%d%d%d", &A, &B, &C);
		++A;
		++B;
		if (!C) Merge(A, B);
		else {
			ns.push_back(make_pair(A, B));
			hav1 = 1;
		}
	}
	for (pair<int, int> p : ns)
		if (same(p.first, p.second)) {
			printf("No\n");
			return 0;
		}
	int block = 0;
	for (int i = 1; i <= N; ++i)
		if (fa[i] == i)
			++block;
	if (M == N - 1) {
		if (!hav1) printf("Yes\n");
		else printf("No\n");
		return 0;
	}
	if (block == 2) {
		if (hav1 == 1) printf("No\n");
		else printf("Yes\n");
		return 0;
	}
	if (block == 1) {
		if (M != N - 1) printf("No\n");
		else printf("Yes\n");
		return 0;
	}
	if (M < N) {
		printf("No\n");
		return 0;
	} else if (M > (N - block) + (ll)block * (block - 1) / 2) {
		printf("No\n");
		return 0;
	} else printf("Yes\n");
	return 0;
}