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;

#define MOD 998244353

int n, q;
long long m;
int par[100005];
vector<pair<int, int>> v;

int find(int a) {
	return par[a] == a ? a : par[a] = find(par[a]);
}

int main() {
	scanf("%d%lld%d", &n, &m, &q);
	for (int i = 0; i < n; i++)
		par[i] = i;
	while (q--) {
		int ta, tb, tc;
		scanf("%d%d%d", &ta, &tb, &tc);
		if (tc)
			v.emplace_back(ta, tb);
		else {
			ta = find(ta);
			tb = find(tb);
			if (ta != tb)
				par[ta] = tb;
		}
	}
	int cnt = 0;
	for (int i = 0; i < n; i++)
		if (find(i) == i)
			cnt++;
	if (cnt <= 2 && !v.empty()) {
		puts("No");
		return 0;
	}
	for (auto& i : v)
		if (par[i.first] == par[i.second]) {
			puts("No");
			return 0;
		}
	long long mi = v.empty() ? n - 1 : n;
	long long ma = (n - 1) - (cnt - 1) + (long long)cnt * (cnt - 1) / 2;
	puts(mi <= m && m <= ma ? "Yes" : "No");
}