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>
void solve(int[], int[], int, int, int, int);
int P;
int main(){
  int n, q, i, j, cnt;
  int B[20], Pb[20];
  scanf("%d", &n);
  int A[n];
//  A = (int *)malloc(sizeof(int) * n);
  for(i = 0; i < n; i++)
    scanf("%d", &A[i]);
  scanf("%d", &q);
  int M[q];
  for(i = 0; i < q; i++){
    scanf("%d", &M[i]);
    P = 0;
    cnt = 0;
    for(j = 0; j < n; j++){
      if(A[j] <= M[i]){
	B[cnt] = A[j];
	Pb[cnt] = 0;
	cnt++;
      }
    }
    solve(B, Pb, 0, cnt, M[i], 0);
    if(P)
      printf("yes\n");
    else
      printf("no\n");
  }
//  free(A);
  return 0;
}

void solve(int B[], int Pb[], int j, int cnt, int M, int d){
  int i;
  for(i = j; i < cnt; i++){
    if(!Pb[i]){
      if(B[i] + d == M){
	P = 1;
	break;
      }
      else if(B[i] + d < M){
	Pb[i] = 1;
	solve(B, Pb, i, cnt, M, d + B[i]);
	Pb[i] = 0;
      }
    }
    if(P)
      break;
  }
}