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:
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#define pii pair<int, int>
#define vii vector<pii>
#define pb push_back
#define fi first
#define se second
using namespace std;
typedef long long ll;
const int MAXN = 1e6 + 5;
int par[MAXN];

inline int qry(int x) { return x == par[x] ? x : par[x] = qry(par[x]); }
inline void join(int u, int v) { par[qry(u)] = qry(v); }

int main() {
	int n, q; ll m;
	scanf("%d %lld %d", &n, &m, &q);
	for (int i = 0; i < n; i++) par[i] = i;
	vii edg;
	int cnt = 0;
	for (int i = 0; i < q; i++) {
		int a, b, c;
		scanf("%d %d %d", &a, &b, &c);
		if (c == 0) {
			if (qry(a) != qry(b))
				cnt++;
			join(a, b);
		} else {
			if (m == n-1)
				return !printf("No\n");
			edg.pb(make_pair(a,b));
		}
	}
	if (m == n-1)
		return !printf("Yes\n");
	for (auto it: edg)
		if (qry(it.fi) == qry(it.se))
			return !printf("No\n");
	ll tmp = 0;
	for (int i = 0; i < n; i++)
		if (qry(i) == i)
			cnt++, tmp++;
	if (cnt <= m && m <= cnt + (tmp*(tmp-3)) / 2 && tmp > 2)
		return !printf("Yes\n");
	printf("No\n");
	return 0;
}