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 <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

typedef long long ll;
const int Maxn = 1e5;

int N, K;
ll M;
vector<pair<int, int> > t;

int fa[Maxn + 5];
int find(int x) {
	if(x == fa[x]) return x;
	return fa[x] = find(fa[x]);
}

int main() {
#ifdef LOACL
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif
	scanf("%d %lld %d", &N, &M, &K);
	for(int i = 1; i <= N; i++)
		fa[i] = i;
	for(int i = 1; i <= K; i++) {
		int u, v, typ;
		scanf("%d %d %d", &u, &v, &typ);
		u++, v++;
		if(typ == 0) {
			int fu = find(u), fv = find(v);
			if(fu != fv) fa[fu] = fv;
		} else t.push_back(make_pair(u, v));
	}
	int cnt = 0;
	for(int i = 1; i <= N; i++)
		if(i == find(i)) cnt++;
	ll mx = 1LL * cnt * (cnt - 1) / 2 + N - cnt;
	if(mx < M) {
		puts("No");
		return 0;
	}
	ll mn = (t.empty() ? N - 1 : N);
	if(mn > M) {
		puts("No");
		return 0;
	}
	for(int i = 0; i < (int)t.size(); i++)
		if(find(t[i].first) == find(t[i].second)) {
			puts("No");
			return 0;
		}
	puts("Yes");
	return 0;
}