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 <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <tuple>
template <typename T> constexpr T my_abs(const T &x) { return x < 0 ? -x : x; }
using ll = long long;
constexpr int MOD = 1e9 + 7;
inline int quick_pow(int a, int n)
{
	int res = 1;
	while (n)
	{
		if (n & 1)
			res = (ll)res * a % MOD;
		a = (ll)a * a % MOD;
		n >>= 1;
	}
	return res;
}
int fa[100005], sz[100005];
int getfa(int u) { return u == fa[u] ? u : fa[u] = getfa(fa[u]); }
std::pair<int, int> seq[100005];
int main()
{
	// freopen("AGC038-D.in", "r", stdin);
	int n, q, fix = 0, cnt = 0, rem = 0, comp = 0;
	ll m;
	scanf("%d%lld%d", &n, &m, &q);
	for (int i = 0; i < n; i++)
	{
		fa[i] = i;
		sz[i] = 1;
	}
	for (int i = 0; i < q; i++)
	{
		int u, v, c;
		scanf("%d%d%d", &u, &v, &c);
		if (!c)
		{
			u = getfa(u);
			v = getfa(v);
			if (u == v)
				continue;
			fa[u] = v;
			sz[v] += sz[u];
		}
		else
			seq[cnt++] = {u, v};
	}
	for (int i = 0; i < n; i++)
	{
		if (i == fa[i])
		{
			rem += sz[i] == 1;
			comp++;
			fix += sz[i] - 1;
		}
	}
	for (int i = 0; i < cnt; i++)
	{
		if (getfa(seq[i].first) == getfa(seq[i].second) ||
				(comp == 2 && !rem))
		{
			puts("No");
			return 0;
		}
	}
	puts(m < fix + comp - !cnt ||
		 m > fix + (ll)comp * (comp - 1) / 2 ?
		 "No" : "Yes");
	return 0;
}