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>
#define N 200010
#define f first
#define int long long
#define s second
using namespace std;
typedef pair<int, int> pii;

int n, m, q, pai[N], peso[N];

int k, nao;

vector<pii> T;

int Find(int x)
{
	if(x == pai[x]) return x;

	return pai[x] = Find(pai[x]);
}

void join(int a, int b)
{
	a = Find(a), b = Find(b);

	if(a == b) return;

	k --;

	if(peso[a] > peso[b]) swap(a, b);

	pai[a] = b;

	if(peso[a] == peso[b]) peso[b] ++;
}

int32_t main()
{
	ios::sync_with_stdio(false); cin.tie(0);

	cin>>n>>m>>q;

	for(int i = 0; i < n; i++) pai[i] = i;

	k = n;

	for(int i = 0; i < q; i++)
	{
		int a, b, c;

		cin>>a>>b>>c;

		if(c == 0)
		{
			if(Find(a) != Find(b))
				join(a, b);
			//else nao = 1;
		}

		else T.push_back({a, b});
	}

	if(nao) cout<<"No\n", exit(0);

	if(m == n - 1 and T.empty()) cout<<"Yes\n", exit(0);

	if(m == n - 1 and !T.empty()) cout<<"No\n", exit(0);

	for(auto w: T)
		if(Find(w.f) == Find(w.s))
		{
			cout<<"No\n";

			exit(0);
		}

	int dhi = (n + k*(k - 1)/2 - k);

	if(m <= dhi) cout<<"Yes\n";

	else cout<<"No\n";
}