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 <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int MAXN = 1e5 + 5;

ll M;
int N, Q;
vector <pair <int, int>> v;
int dad[MAXN], rnk[MAXN];
 
int find(int x) {
	return dad[x] == x ? x : dad[x] = find(dad[x]);
}

void join(int x, int y) {
	x = find(x);
	y = find(y);
	if (x == y)
		return;
	if (rnk[x] > rnk[y])
		swap(x, y);
	dad[x] = y;
	rnk[y] += rnk[x] == rnk[y];
}

void load() {
	scanf("%d%lld%d", &N, &M, &Q);
	for (int i = 0; i < N; i++) 
		dad[i] = i;
	while (Q--) {
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		if (c)
			v.push_back({a, b});
		else
			join(a, b);
	}
}

bool solve() {
	if (M == N - 1)
		return v.empty();
	
	for (auto it : v)
		if (find(it.first) == find(it.second))
			return false;
	
	int comps = 0;
	for (int i = 0; i < N; i++)
		comps += find(i) == i;
	
	return comps >= 3 && M <= N + (ll)comps * (comps - 3) / 2;
}

int main() {
	load();
	puts(solve() ? "Yes" : "No");
	return 0;
}