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;
import java.util.Arrays;

public class Main {
	
		static int n;
		static int[] A;
		static int q;
		static int m;
		
		static int[][] dp;
	
	public static void main(String[] args) throws NumberFormatException, IOException {
		// TODO 自動生成されたメソッド・スタブ
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] str;
		
		n = Integer.parseInt(br.readLine());
		A = new int[n];
		
		str = br.readLine().split(" ");
		for(int i = 0; i < n; i++){
			A[i] = Integer.parseInt(str[i]);
		}
		
		q = Integer.parseInt(br.readLine());
		
		str = br.readLine().split(" ");
		for(int i = 0; i < q; i++){
			m = Integer.parseInt(str[i]);
			
			dp = new int[n][m + 2000];
			for(int[] array : dp){
				Arrays.fill(array, -1);
			}
			
			System.out.println(dfs(0, 0) == 1?"yes" : "no");
		}
		
	}
	
	public static int dfs(int i, int sum){
		
		if(sum == m) return 1;
		
		else if(i == n) return 0;
		
		else if(dp[i][sum] != -1) return dp[i][sum];
		
		else if(dfs(i + 1, sum) == 1) return dp[i][sum] = 1;
		
		else if(dfs(i + 1, sum + A[i]) == 1) return dp[i][sum] = 1;
		
		else return dp[i][sum] = 0;
	}
}