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<algorithm>
#include<vector>
using namespace std;
int n, Q, vis[101000], cnt, ct;
long long m;
vector<int>E[101000], G;
struct point {
	int a, b, c;
}w[101000];
void DFS(int a) {
	vis[a] = cnt;
	ct++;
	for (auto &x : E[a]) {
		if (!vis[x])DFS(x);
	}
}
int main() {
	int i;
	scanf("%d%lld%d", &n, &m, &Q);
	for (i = 1; i <= Q; i++) {
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		a++, b++;
		w[i] = { a,b,c };
		if (c == 0) {
			E[a].push_back(b);
			E[b].push_back(a);
		}
	}
	for (i = 1; i <= n; i++) {
		if (!vis[i]) {
			cnt++;
			ct = 0;
			DFS(i);
			G.push_back(ct);
		}
	}
	int chk = 0;
	for (i = 1; i <= Q; i++) {
		if (w[i].c && vis[w[i].a] == vis[w[i].b]) {
			puts("No");
			return 0;
		}
		if (w[i].c)chk = 1;
	}
	if (G.size() == 2) {
		G.clear();
		G.push_back(n);
	}
	long long Mn = n - 1, Mx = 0;
	if (chk)Mn = n;
	for (auto &t : G) {
		Mx += t - 1;
	}
	long long c = G.size();
	Mx += c * (c - 1) / 2;
	if (Mn <= m && m <= Mx) {
		puts("Yes");
	}
	else puts("No");
}