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: Runtime Error)

#include <stdio.h>

#define N 2000
#define M 200

int solve(int);
void way1(int);
void way2(int);

int A[N];
int n;

int main(){

  int ans,q,i;
  int mi[M];

  scanf("%d",&n);

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

  scanf("%d",&q);
  for(i=0 ; i<q ; i++){
    scanf("%d",&mi[i]);
  }


  for(i=0 ; i<q ; i++){
    printf("%d",i);
    ans = solve(A[i]);
    if(ans==1) printf("yes\n");
    else printf("no\n");
  }

  return 0;
}

int solve(int m){

  int n,a,b;

  n==m;

  while(1){
    if(m==0){
      return 1;
      break;
    }
    
    else if(m<n){
      return 0;
      break;
    }

    else{
      way1(m);
      way2(m);
    }
  }
}


void way1(int m){

  static int j=0;

  solve(m-A[j]);

  j++;

  if(j==n) j=0;
}
    

void way2(int m){

  solve(m);

}