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;

using ll = long long;

struct UF {
    int n;
    vector<int> par;
    UF(int n) : n(n) {
        for(int i = 0; i < n; i++) par.push_back(i);
    }
    int find(int a){
        if(a != par[a]) par[a] = find(par[a]);
        return par[a];
    }
    void join(int a, int b){
        par[find(a)] = find(b);
    }
};

int main(){
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int n;
	ll m;
	int q;
	cin >> n >> m >> q;
	UF uf(n);
	vector<pair<int,int> > bp;
	for(int i = 0; i < q; i++){
		int a, b, c;
		cin >> a >> b >> c;
		if(c == 0){
			uf.join(a, b);
		} else {
			bp.push_back({a, b});
		}
	}
	for(pair<int,int> x : bp){
		if(uf.find(x.first) == uf.find(x.second)){
			cout << "No" << '\n';
			return 0;
		}
	}
	if(m == n-1){
		if(bp.empty()){
			cout << "Yes" << '\n';
		} else {
			cout << "No" << '\n';
		}
		return 0;
	}
	ll ncc = 0;
	for(int i = 0; i < n; i++){
		if(uf.find(i) == i) ncc++;
	}
	ll g = n-1 + ncc * (ncc - 1) / 2 - (ncc - 1);
	if(m <= g){
		cout << "Yes" << '\n';
	} else {
		cout << "No" << '\n';
	}
	return 0;
}