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>

typedef long long LL;
const int N = 100005;
int n, q, f[N], c; LL m;
std::vector<std::pair<int, int>> v1, v2;
int find(int x) { return x == f[x] ? x : f[x] = find(f[x]); }
void link(int x, int y) { f[find(x)] = find(y); }
int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n >> m >> q;
	for (int i = 0; i < n; ++i) f[i] = i;
	for (int i = 1, x, y, z; i <= q; ++i)
		std::cin >> x >> y >> z, (z ? v2 : v1).emplace_back(x, y);
	for (auto e : v1) link(e.first, e.second);
	for (auto e : v2) if (find(e.first) == find(e.second)) return std::cout << "No\n", 0;
	for (int i = 0; i < n; ++i) c += f[i] == i;
	std::cout << ((v2.size() ? std::max(3, c) : c - 1) <= m - (n - c) && m - (n - c) <= (LL) c * (c - 1) / 2 ? "Yes" : "No") << '\n';
	return 0;
}