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 <stdlib.h>

void makec(int);
void rec(int);
void sold(void);

//組み合わせの配列生成まではできた
//kumiawase no array made ha dekita

//m 最初のDataArrayの個数
//n 二つめのたしてくやつの個数
//qa[] dataarray
//s[] たしてくやつ
//ans[] s[]の組み合わせ
//qaa[]

int m,n,s[200],ans[20],qa[20],de[200];

int main(){
  int i,k;

  //for(i=0;i<20;i++)qaa[i]=0;

  scanf("%d",&m);
  for(i=0;i<m;i++)scanf("%d",&qa[i]);

  scanf("%d",&n);
  for(i=0;i<n;i++)scanf("%d",&s[i]);
  
  makec(n);

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

  //qaa[i]==1/other
  //i行目にyes/no
  return 0;
}


void makec(int n){
  int i;
  for(i=0;i<m;i++)ans[i] = 0;
  for(i=0;i<n;i++)de[i] = 0;
  
  rec(0);
}

//組み合わせの0,1をつくる
void rec(int i){
  int kaunto;
  if(i==m){
    //for(kaunto=0;kaunto<m;kaunto++){
      //printf("%d ",ans[kaunto]);
      //sold();
    //}
    sold();
    //printf("\n");
  }else{
    rec(i+1);
    ans[i] = 1;
    rec(i+1);
    ans[i] = 0;
  }
}

void sold(void){
  int i,cal=0;
  for(i=0;i<m;i++){
    if(ans[i]==1){
      cal+=qa[i];
    }
  }
  //printf("#%d#\n",cal);
  for(i=0;i<n;i++){
    if(cal==s[i]){
      //printf("yes");
      // printf("%d is s[%d]!!\n",cal,i);
      de[i]=1;
    }
  }
}