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;
const int N = 100005;
int n, q;
long long m;
int fa[N];
int Find(int x) {return fa[x] == x ? x : fa[x] = Find(fa[x]);}
void Union(int x, int y) {
	x = Find(x), y = Find(y);
	if (x != y) fa[x] = y;
}
int main() {
	scanf("%d%lld%d", &n, &m, &q);
	vector<pair<int, int> > vec;
	for (int i = 0; i < n; i++) fa[i] = i;
	while (q--) {
		int u, v, c;
		scanf("%d%d%d", &u, &v, &c);
		if (c == 0)
			Union(u, v);
		else
			vec.push_back(make_pair(u, v));
	}
	if (m == n - 1) {
		if (vec.size()) puts("No");
		else puts("Yes");
		return 0;
	}
	for (pair<int, int> p : vec) {
		if (Find(p.first) == Find(p.second)) {
			puts("No");
			return 0;
		}
	}
	int cnt = 0;
	for (int i = 0; i < n; i++) if (fa[i] == i) cnt++;
	if (m > n - cnt + 1ll * (cnt - 1) * cnt / 2) puts("No");
	else puts("Yes");
	return 0;
}