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: Time Limit Exceeded)

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

void bb(int *b, int n){
  int i;
  for(i = 0; i < n; i++){
    if(b[i] == 0){
      b[i] = 1;
      break;
    }
    else{
      b[i] = 0;
    }
  }
}

int roopNum(int n){
  int i;
  int sum = 0;
  for(i = 0; i < n; i++){
    sum += pow(2, i);
  }
  return sum;
}

void dis(int *b, int n){
  int i;
  for(i = 0; i < n; i++){
    printf("%d",b[i]);
  }
  printf("\n");
}

void Nacht(int *A, int n, int m, int r){
  int *b;
  int i, j, sum;
  b = (int *)malloc(sizeof(int)*n);
  for(i =0; i < n; i++){
      b[i] = 0;
  }
  for(i = 0; i < r; i++){
    bb(b, n);
    //dis(b, n);
    sum = 0;
    for(j = 0; j < n; j++){
      if(b[j] == 1) sum += A[j];
    }
    if(sum == m){
      printf("yes\n");
      return;
    }
  }
  printf("no\n");
}

int Makem(int *A, int i,int n, int m){
}

int main(){

  int n, q, *A, *m, i, j, N, v,k, cnt;
  int r;

  //入力

  scanf("%d",&n);
  A = (int *)malloc(sizeof(int)*n);

  for(i=0; i<n; i++){
    scanf("%d",&A[i]);
  }
  scanf("%d",&q);
  m = (int *)malloc(sizeof(int)*q);

  for(i=0; i<q; i++){
    scanf("%d",&m[i]);
  }
   r = roopNum(n);
  for(i=0; i<q; i++){
    Nacht(A, n, m[i], r);
  }
  return 0;
}