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.util.NoSuchElementException;
 class FS {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        } else {
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }

    private int readByte() {
        return hasNextByte() ? buffer[ptr++] : -1;
    }

    private boolean isPrintableChar(int c) {
        return 33 <= c && c <= 126;
    }

    private void skipUnprintable() {
        while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
    }

    public boolean hasNext() {
        skipUnprintable();
        return hasNextByte();
    }

    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while (isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    public int nextInt() {
        return (int) nextLong();
    }

    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while (true) {
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            } else if (b == -1 || !isPrintableChar(b)) {
                return minus ? -n : n;
            } else {
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
}
public class Main {
    static int[] a;
    static int[][] dp;

    public static void main(String[] args) {
        StringBuilder sb = new StringBuilder();
        FS fs = new FS();
        int n = fs.nextInt();
        a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = fs.nextInt();
        }
        int q = fs.nextInt();
        //    qu = new int[q];
        dp = new int[2000][2000];
        boolean j = true;
        for (int i = 0; i < q; i++) {
            int t = fs.nextInt();
            if (solve(0, t) == 1) {
                sb.append("yes\n");
            } else {
                sb.append("no\n");
            }
        }
        System.out.print(sb.toString());
    }

    private static int solve(int p, int t) {
       if (p<0||t<0)return -1;
        if (dp[p][t] != 0) return dp[p][t];
        if (t == 0) dp[p][t] = 1;

        else if (p >= a.length) dp[p][t] = -1;
        else if (solve(p + 1, t - a[p]) == 1 || solve(p + 1, t) == 1) {
            dp[p][t] = 1;
        } else dp[p][t] = -1;
        return solve(p, t);
    }
}