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;
typedef bool boolean;

const int N = 1e5 + 5;

#define pii pair<int, int>
#define ll long long

int n, q;
ll m;
int uf[N];
vector<pii> E0, E1;

int find(int x) {
	return uf[x] == x ? x : uf[x] = find(uf[x]);
}
boolean unit(int x, int y) {
	x = find(x), y = find(y);
	if (x ^ y) {
		uf[x] = y;
		return true;
	}
	return false;
}

void quitf(boolean expression) {
	if (expression) {
		puts("No");
		exit(0);
	}
}

int main() {
	scanf("%d%lld%d", &n, &m, &q);
	for (int i = 1; i <= n; i++)
		uf[i] = i;
	for (int i = 1, u, v, opt; i <= q; i++) {
		scanf("%d%d%d", &u, &v, &opt);
		++u, ++v;
		if (opt == 0) {
			E0.emplace_back(u, v);
		} else {
			E1.emplace_back(v, u);
		}
	}
	for (auto e : E0)
		unit(e.first, e.second);
	for (auto e : E1)
		quitf(find(e.first) == find(e.second));
	int comp = 0;
	for (int i = 1; i <= n; i++)
		comp += find(i) == i;
	quitf(E1.size() && comp == 2);
	ll mi = ((E1.size()) ? (n) : (n - 1));
	ll mx = (n - comp) + (comp * 1ll * (comp - 1) >> 1);
	quitf(m < mi || m > mx);
	puts("Yes");
	return 0;
}