Problem:
Exhaustive Search

Write a program which reads a sequence A of n elements and an integer M, and outputs "yes" if you can make M by adding elements in A, otherwise "no". You can use an element only once.

You are given the sequence A and q questions where each question contains Mi.

Input

In the first line n is given. In the second line, n integers are given. In the third line q is given. Then, in the fourth line, q integers (Mi) are given.

Output

For each question Mi, print yes or no.

Constraints

n ≤ 20

q ≤ 200

1 ≤ elements in A ≤ 2000

1 ≤ Mi ≤ 2000

Sample Input 1

5
1 5 7 10 21
8
2 4 17 8 22 21 100 35

Sample Output 1

no
no
yes
yes
yes
yes
no
no

Notes

You can solve this problem by a Burte Force approach. Suppose solve(p, t) is a function which checkes whether you can make t by selecting elements after p-th element (inclusive). Then you can recursively call the following functions:

solve(0, M)

solve(1, M-{sum created from elements before 1st element})

solve(2, M-{sum created from elements before 2nd element})

...

The recursive function has two choices: you selected p-th element and not. So, you can check solve(p+1, t-A[p]) and solve(p+1, t) in solve(p, t) to check the all combinations.

For example, the following figure shows that 8 can be made by A[0] + A[2].

Solution in JAVA (judge verdict: Accepted)

import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.TreeSet;

 
public class Main {
  public static void main(String[] args) {
    InputReader sc = new InputReader(System.in);
 
    int n = sc.nextInt();
    int[] a = new int[n];
    for(int i = 0; i < n; i++){
      a[i] = sc.nextInt();
    }
    int q = sc.nextInt();

    boolean[] hantei = new boolean[2001];
    hantei[0] = true;
    for(int i = 1; i < 2001; i++){
      hantei[i] = false;
    }

    for(int i = 0; i < n; i++){
      int nownum = a[i];
      for(int j = 2000 - nownum; j >= 0; j--){
        hantei[j + nownum] |= hantei[j];
      }
    }

    for(int i = 0; i < q; i++){
      System.out.println(hantei[sc.nextInt()] ? "yes" : "no");
    }
  }
 
  
// LinkedList<Integer>[] setsu = new LinkedList[n];
// for(int i = 0; i < n; i++){
//   setsu[i] = new LinkedList<Integer>();
// } 

 //ここからテンプレ
  static class InputReader {
      private InputStream stream;
      private byte[] buf = new byte[1024];
      private int curChar;
      private int numChars;
      private SpaceCharFilter filter;

      public InputReader(InputStream stream) {
          this.stream = stream;
      }
 
      public int next() {
          if (numChars == -1)
              throw new InputMismatchException();
          if (curChar >= numChars) {
              curChar = 0;
              try {
                  numChars = stream.read(buf);
              } catch (IOException e) {
                  throw new InputMismatchException();
              }
              if (numChars <= 0)
                  return -1;
          }
          return buf[curChar++];
      }
 
      public String nextStr() {
        int c = next();
        while(isSpaceChar(c)){c = next();}
        StringBuffer str = new StringBuffer();
        do{
          str.append((char)c);
          c = next();
        }while(!isSpaceChar(c));
        return str.toString();
      }

      public int nextInt() {
          int c = next();
          while (isSpaceChar(c))
              c = next();
          int sgn = 1;
          if (c == '-') {
              sgn = -1;
              c = next();
          }
          int res = 0;
          do {
              if (c < '0' || c > '9')
                  throw new InputMismatchException();
              res *= 10;
              res += c - '0';
              c = next();
          } while (!isSpaceChar(c));
          return res * sgn;
      }
 
      public boolean isSpaceChar(int c) {
          if (filter != null)
              return filter.isSpaceChar(c);
          return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
      }
 
      public interface SpaceCharFilter {
          public boolean isSpaceChar(int ch);
      }
  }
}