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 MAXN = 100005;
int n, q;
long long m;
vector<pair<int, int> > con[2];
int p[MAXN];
int find(int i) {
	if (p[i] == -1 || p[i] == i) return i;
	return p[i] = find(p[i]);
}
bool un(int u, int v) {
	u = find(u);
	v = find(v);
	if (u == v) return false;
	p[u] = v;
	return true;
}
int main() {
	memset(p, -1, sizeof(p));
	scanf("%d %lld %d", &n, &m, &q);
	bool tree = true;
	for (int i = 0; i < q; i++) {
		int a, b, c;
		scanf("%d %d %d", &a, &b, &c);
		con[c].push_back({a, b});
		if (c) tree = false;
	}
	for (int i = 0; i < con[0].size(); i++) {
		un(con[0][i].first, con[0][i].second);
	}
	for (int i = 0; i < con[1].size(); i++) {
		if (find(con[1][i].first) == find(con[1][i].second)) {
			printf("No\n");
			return 0;
		}
	}
	int comp = 0;
	for (int i = 0; i < n; i++) if (find(i) == i) comp++;
	long long minM = n-1;
	if (!tree) {
		minM = n;
		if (comp <= 2) {
			printf("No\n");
			return 0;
		}
	}
	long long maxM = n-comp + (comp * 1ll * (comp - 1)) / 2;
	if (minM <= m && m <= maxM) {
		printf("Yes\n");
	} else printf("No\n");
}