import java.util.*;


public class BinPacking
{

   /* -----------------------------------------------------------
      SolveBP(k1, s1, k2, s2, k3, s3, res, Cap):

          returns the min # bins needed to pack:

		k1 items of size s1
		k2 items of size s2
		k3 items of size s3

	  using bins of capacity Cap when the
	  current bin has a residual capacity of "res"
      ----------------------------------------------------------- */
   public static int solveBP(int k1, int s1, int k2, int s2, int k3, int s3,
			     int res, int Cap)
   {
      // Write you (recursive) solution here

   }


   public static void main(String[] args)
   {
      int n1, n2, n3;
      int s1, s2, s3;
      int Cap;
      int out;

      Scanner in = new Scanner(System.in);

      System.out.print("Size of item 1   = ");
      s1 = in.nextInt();
      System.out.print("Number of item 1 = ");
      n1 = in.nextInt();
      System.out.print("Size of item 2   = ");
      s2 = in.nextInt();
      System.out.print("Number of item 2 = ");
      n2 = in.nextInt();
      System.out.print("Size of item 3   = ");
      s3 = in.nextInt();
      System.out.print("Number of item 3 = ");
      n3 = in.nextInt();

      System.out.println();
      System.out.print("Capacity of the bin = ");
      Cap = in.nextInt();

      if ( s1 > Cap || s2 > Cap || s3 > Cap )
      {
         System.out.println("Input error: some item's size is too large for bin");
         System.exit(1);
      }
	 
      out = solveBP(n1, s1, n2, s2, n3, s3, Cap, Cap);

      System.out.println("Min # bins needed = " + out);
   }

}