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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

  public static void main(String[] args) throws IOException {
    
    try(BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

      StrSplitToInt str;
      int n = Integer.parseInt(br.readLine());
      int[] seqA = new int[n];
      str = new StrSplitToInt(br.readLine(), " ");
      for(int i=0; i<n; i++) {
        seqA[i] = str.nextInt();
      }

      int q = Integer.parseInt(br.readLine());
      str = new StrSplitToInt(br.readLine(), " ");
      for(int i=0; i<q; i++) {
        if( solve(n, seqA, 0, str.nextInt()) )
          System.out.println("yes");
        else
          System.out.println("no");
      }
    }
  }

  private static boolean solve(int n, int[] a, int i, int m) {
    if( m == 0 ) return true;
    if( i >= n ) return false;

    boolean res = solve(n, a, i+1, m) || solve(n, a, i+1, m-a[i]);

    return res;
  }
}
 
class StrSplitToInt {
  final String s;
  final String split;
  final int endPos;
  final int splitLen;
  int currentPos;
  int nextPos;
 
  public StrSplitToInt(String s,String split) {
    this.s = s;
    this.split = split;
    currentPos = 0;
    nextPos = s.indexOf(split, currentPos);
    endPos = s.length();
    splitLen = split.length();
  }
 
  public int nextInt() {
    if( currentPos < endPos ) {
      int ret = Integer.parseInt( s.substring(currentPos, nextPos) );
      currentPos = nextPos + splitLen;
      nextPos = s.indexOf(split, currentPos);
      if(nextPos == -1) nextPos = endPos;
 
      return ret;
    }
    else
      return -1;
  }
}