#include <iostream>
#include <vector>
#include <numeric>
using namespace std;
typedef long long int ll;

vector<int> G[100100];
int used[100100];

struct UnionFind{
    vector<int> par,num;
    vector<bool> done;
    UnionFind(int n):par(n),num(n,1),done(n,false){
        iota(par.begin(),par.end(),0);  //include<numeric>
    }
    int find(int v){
        return (par[v]==v)?v:(par[v]=find(par[v]));
    }
    void unite(int u,int v){
        u=find(u),v=find(v);
        if(u==v)return;
        if(num[u]<num[v])swap(u,v);
        num[u]+=num[v];
        par[v]=u;
        done[u]=done[u]|done[v];
    }
    bool same(int u,int v){
        return find(u) == find(v);
    }
    bool ispar(int v){
        return v=find(v);
    }
    int size(int v){
        return num[find(v)];
    }
};

int main(){
	ll n,m,q; cin >> n >> m >> q;
	UnionFind uf(n);
	vector<pair<int,int> > v;
	for(int i=0;i<q;i++){
		int a,b,c; cin >> a >> b >> c;
		if(c==0){
			uf.unite(a,b);
		}
		else{
			v.push_back(make_pair(a,b));
		}
	}
	for(auto p:v){
		if(m==n-1||uf.same(p.first,p.second)){
			cout << "No" << endl;
			return 0;
		}
	}
	if(m==n-1){
		cout << "Yes" << endl;
		return 0;
	}
	ll cnt=0;
	for(int i=0;i<n;i++){
		if(uf.find(i)==i)cnt++;
	}
	if(2*m<=2*n+cnt*(cnt-3)){
		cout << "Yes" << endl;
	}
	else{
		cout << "No" << endl;
	}
}
### Prompt

Your task is to create a CPP solution to the following 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

### Response

```cpp
#include <iostream>
#include <vector>
#include <numeric>
using namespace std;
typedef long long int ll;

vector<int> G[100100];
int used[100100];

struct UnionFind{
    vector<int> par,num;
    vector<bool> done;
    UnionFind(int n):par(n),num(n,1),done(n,false){
        iota(par.begin(),par.end(),0);  //include<numeric>
    }
    int find(int v){
        return (par[v]==v)?v:(par[v]=find(par[v]));
    }
    void unite(int u,int v){
        u=find(u),v=find(v);
        if(u==v)return;
        if(num[u]<num[v])swap(u,v);
        num[u]+=num[v];
        par[v]=u;
        done[u]=done[u]|done[v];
    }
    bool same(int u,int v){
        return find(u) == find(v);
    }
    bool ispar(int v){
        return v=find(v);
    }
    int size(int v){
        return num[find(v)];
    }
};

int main(){
	ll n,m,q; cin >> n >> m >> q;
	UnionFind uf(n);
	vector<pair<int,int> > v;
	for(int i=0;i<q;i++){
		int a,b,c; cin >> a >> b >> c;
		if(c==0){
			uf.unite(a,b);
		}
		else{
			v.push_back(make_pair(a,b));
		}
	}
	for(auto p:v){
		if(m==n-1||uf.same(p.first,p.second)){
			cout << "No" << endl;
			return 0;
		}
	}
	if(m==n-1){
		cout << "Yes" << endl;
		return 0;
	}
	ll cnt=0;
	for(int i=0;i<n;i++){
		if(uf.find(i)==i)cnt++;
	}
	if(2*m<=2*n+cnt*(cnt-3)){
		cout << "Yes" << endl;
	}
	else{
		cout << "No" << endl;
	}
}
```