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 C (judge verdict: Accepted)

#include <stdio.h>
#include <math.h>
#include <memory.h>

char cache[2000+1];

inline void create_cache(int A[], int n, int flag) {
	int i, s;
	for(i=0,s=0;i<n;i++) {
		if((flag&(1<<i)) != 0) {
			s += A[i];
		}
	}
	cache[s] = 1;
}

int main(void) {
	int i, j, n, q, size, t, f;
	int A[20];
	int M[200];
	memset(cache, 0x00, sizeof(cache));

	scanf("%d\n", &n);
	for(i=0;i<n;i++) {
		scanf("%d", &A[i]);
	}
	size=pow(2, n);
	for(i=0;i<size;i++) {
		create_cache(A, n, i);
	}

	scanf("%d\n", &q);
	for(i=0;i<q;i++) {
		scanf("%d", &M[i]);
		if(cache[M[i]]) {
			printf("yes\n");
		} else {
			printf("no\n");
		}
	}
	
	return 0;
}