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 <cstdio>
#include <queue>
using namespace std;

int main()
{
  int n,q;
  long long m;
  scanf("%d%lld%d",&n,&m,&q);
  vector<vector<int> > e(n);
  vector<int> A,B;
  for(int i=0;i<q;i++){
    int a,b,c;
    scanf("%d%d%d",&a,&b,&c);
    if(c==0){
      e[a].push_back(b);
      e[b].push_back(a);
    }
    else{
      A.push_back(a);
      B.push_back(b);
    }
  }
  bool ans;
  if(m==n-1){
    ans=(A.size()==0);
  }
  else{
    vector<int> P(n,-1);
    int k=0;
    for(int r=0;r<n;r++){
      if(P[r]!=-1)
	continue;
      k++;
      queue<int> Q;
      Q.push(r);
      P[r]=r;
      while(!Q.empty()){
	int i=Q.front();
	Q.pop();
	for(int x=0;x<e[i].size();x++){
	  int j=e[i][x];
	  if(P[j]==-1){
	    P[j]=r;
	    Q.push(j);
	  }
	}
      }
    }
    ans=(m<=(n-k)+(long long)k*(k-1)/2);
    for(int t=0;t<A.size();t++){
      if(P[A[t]]==P[B[t]]){
	ans=false;
      }
    }
  }
  puts(ans?"Yes":"No");
  return 0;
}