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>
#define debug(x) cerr << #x << " " << (x) << endl
using namespace std;
const int N = 100005, Q = 100005;
int n, q, a[Q], b[Q], c[Q], fa[N];
long long m;
bool vis[N], flag = true;
int findroot (int u) {
	return fa[u] == u ? u : fa[u] = findroot(fa[u]);
}
int main () {
	scanf("%d%lld%d", &n, &m, &q);
	for (int i = 1; i <= n; i++) vis[i] = false;
	for (int i = 1; i <= q; i++) scanf("%d%d%d", &a[i], &b[i], &c[i]), a[i]++, b[i]++;
	for (int i = 1; i <= n; i++) fa[i] = i;
	for (int i = 1; i <= q; i++) {
		if (!c[i]) {
			vis[a[i]] = vis[b[i]] = true;
			fa[findroot(a[i])] = findroot(b[i]);
		}
	}
	if (m == n - 1) {
		for (int i = 1; i <= q; i++) {
			if (c[i]) flag = false;
		}
		printf(flag ? "Yes\n" : "No\n");
	}
	else {
		int comp = 0;
		for (int i = 1; i <= n; i++) {
			if (fa[i] == i) comp++;
		}
		if (n - comp + 1ll * comp * (comp - 1) / 2 < m) flag = false;
		for (int i = 1; i <= q; i++) {
			if (c[i] && findroot(a[i]) == findroot(b[i])) flag = false;
		}
		printf(flag ? "Yes\n" : "No\n");
	}
	return 0;
}