// This is to solve the exact knapsack pproblem based on
// Algorithm 2. Array x is to hold binary sequences.
// x[i]=1 means item i is chosen, x[i]=0, i not chosen.
// The exact knapsack is to choose some a[i]'s so that
// the sum of those can be equal to b.

int b, i,n, s, t;
int a[100],d[100],up[100], w[100], x[100];
out(int a[])
{
int i;
  for(i=1; i<=n; i++)printf("%d ", a[i]);
  printf("\n");
}
main(){
  scanf("%d", &n);
  scanf("%d", &b);
  for(i=1;i<=n;i++) x[i]=0;
  for(i=1;i<=n;i++) a[i]=random()%100000;
  for(i=1;i<=n;i++) w[i]=random()%100;
  out(a);
  for(i=0;i<=n;i++) up[i]=i;
  s=0;
  t=clock();
  do{
//    out(x);
    i=0;
    while(x[i+1]==1){ 
      x[i+1]=0;
      s=s-a[i+1];
      i=i+1;
    }
    x[i+1]=1;
    s=s+a[i+1];
    if(s==b){out(x); break;}
  }
  while (i!=n);
  printf("time %d\n", (clock()-t)/1000);
}