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 <iostream>
#include <vector>
using namespace std;
typedef long long ll;
class UnionFind{
private:
    vector<int> p,s;
	int cnt;
public:
	UnionFind(){}
	UnionFind(int N){
		cnt = N;
		p = s = vector<int>(N+1,0);
		for(int i=1;i<=N;i++){
			p[i] = i; s[i] = 1;
		}
	}
	int find(int x){
		if(p[x]==x) return x;
		else return p[x] = find(p[x]);
	}
	void unite(int x,int y){
		x = find(x); y = find(y);
		if(x==y) return;
		if(s[x]>s[y]){
			p[y] = x;
			s[x] += s[y];
		}else{
			p[x] = y;
			s[y] += s[x];
		}
		cnt--;
	}
	bool is_same_set(int x,int y) {return find(x)==find(y);}
	ll size(int x) {return s[find(x)];}
	int compnents_number(){return cnt;}
};
   
int main(){
    ll N,M,Q;
    cin >> N >> M >> Q;
    vector<pair<int,int>> v[2];
    for(int i=0;i<Q;i++){
        int a,b,c;
        cin >> a >> b >> c;
        a++; b++;
        v[c].push_back({a,b});
    }
    if(M==N-1){
        cout << (v[1].empty()? "Yes":"No") << endl;
        return 0;
    }
    UnionFind uf(N);
    for(auto x:v[0]){
        uf.unite(x.first,x.second);
    }
    bool ok = true;
    for(auto x:v[1]){
        if(uf.is_same_set(x.first,x.second)) ok = false;
    }
    ll si = uf.compnents_number();
    ll upp = N+si*(si-3)/2;
    ok &= (M<=upp);
    cout << (ok? "Yes":"No") << endl;
}