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>

int solve(int i, int a[], int m, int n) {
  int res;
  if (m == 0) {
    return 1;
  }
  if (i >= n) {
    return 0;
  }
  res = solve(i + 1, a, m, n) || solve(i + 1, a, m - a[i], n);
  return res;
}

int main(void) {
  int i, j, a[10000], m[10000], n, l1, l2;
  char buf[2];
  scanf("%d", &l1);
  for (n = 0; n < l1 && (j = scanf("%d%1[\n]", a+n, buf)) == 1; n++) ;
  if (j == 2)
  n++;
  scanf("%d", &l2);
  for (n = 0; n < l2 && (j = scanf("%d%1[\n]", m+n, buf)) == 1; n++) ;
  if (j == 2)
  n++;

  for (i = 0; i < l2; i++) {
    if (solve(0, a, m[i], l1) == 1) {
      printf("yes\n");
    } else {
      printf("no\n");
    }
  }
  return 0;
}