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 rep(i, n) for(int i =0; i < n; i++)
using namespace std;

long long n, m, q;
vector< pair<int, int> > lmt;
int par[100100];
bool used[100100];
int get_par(int a){
	return (par[a] = (par[a] == a) ? a : get_par(par[a]));
}
void unite(int a, int b){
	if(get_par(a) != get_par(b)) par[get_par(a)] = get_par(b);
}

int main(){
	ios::sync_with_stdio(false);
	cin >> n >> m >> q;
	rep(i, n) par[i] = i;
	rep(i, q){
		int a, b, c;
		cin >> a >> b >> c;
		if(c == 0) unite(a, b);
		else lmt.push_back(make_pair(a, b));
	}
	if(m == n-1){
		if(lmt.size() == 0){
			cout << "Yes" << endl;
		} else cout << "No" << endl;
		return 0;
	}
	rep(i, (int)lmt.size()){
		if(get_par(lmt[i].first) == get_par(lmt[i].second)){
			cout << "No" << endl;
			return 0;
		}
	}
	long long cnt = 0;
	rep(i, n) if(!used[get_par(i)]) cnt++, used[get_par(i)] = 1;
	if(cnt <= 2){
		cout << "No" << endl;
		return 0;
	}
	if(m > 0LL + n + 1LL * cnt * (cnt - 1LL) / 2LL - cnt) cout << "No" << endl;
	else cout << "Yes" << endl;
	return 0;
}