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 {
		
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		
		int n = Integer.parseInt(reader.readLine());
		String[] inputs = reader.readLine().split(" ");
		int[] arr = new int[n];
		for (int i = 0; i < n; i++) {
			arr[i] = Integer.parseInt(inputs[i]);
		}
		
		int q = Integer.parseInt(reader.readLine());
		inputs = reader.readLine().split(" ");
		for (int i = 0; i < q; i++) {
			boolean canMakeNumber = checkIfCanMakeNumber(arr, Integer.parseInt(inputs[i]));
			if (canMakeNumber) {
				System.out.println("yes");
			} else {
				System.out.println("no");
			}
		}
	}

	private static boolean checkIfCanMakeNumber(int[] arr, int target) {		
		return checkIfCanMakeNumber(0, arr, target);
	}
	
	private static boolean checkIfCanMakeNumber(int idx, int[] arr, int target) {
		
		if (target == 0) {
			return true;
		}
		if (idx >= arr.length) {
			return false;
		}
		
		boolean ret1 = checkIfCanMakeNumber(idx + 1, arr, target);
		boolean ret2 = checkIfCanMakeNumber(idx + 1, arr, target - arr[idx]);
		
		if (ret1 == true || ret2 == true) {
			return true;
		}
		return false;
	}
	
	/*
	private static boolean checkIfCanMakeNumber(int[] arr, int target) {
		return checkIfCanMakeNumber(null, arr, target);
	}

	private static boolean checkIfCanMakeNumber(int[] base, int[] arr, int target) {

		boolean ret = false;		
		if (base != null && base.length > arr.length) {
			return ret;
		}

		int startPos = 0;
		int baseLen = 0;
		if (base != null) {
			baseLen = base.length;
			startPos = base[baseLen - 1] + 1;
		}

		int[] newBase = new int[baseLen + 1];
		for (int i = startPos; i < arr.length; i++) {
			int sum = 0;
			for (int j = 0; j < baseLen; j++) {
				sum += arr[base[j]];
				if (sum == target) {
					
					return true;
				}
				if (sum > target) {
					return false;
				}
				newBase[j] = base[j];
			}
			sum += arr[i];
			if (sum == target) {
				return true;
			}

			newBase[baseLen] = i;
			ret = checkIfCanMakeNumber(newBase, arr, target);
			if (ret == true) {
				return true;
			}
		}
		return ret;
	}
	*/
}