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 printArray(int array[], int n) {
    for (int i = 0; i < n; i++) {
        if (i != 0) printf(" ");
        printf("%3d", array[i]);
    };
}
void printArrayNoSpace(int array[], int n) {
    for (int i = 0; i < n; i++) {
        if (i != 0) printf(" ");
        printf("%d", array[i]);
    };
}
void printBooleanArray(int array[], int n) {
    for (int i = 0; i < n; i++) {
        if (i != 0) printf("\n");
        if (array[i])
            printf("yes");
        else
            printf("no");
    };
}

// can a value be expressed by the sum of integers (in distinct index) in the array?
int expressibleBySumOf( int value,      // t - value to be expressed
                        int hereIndex,  // p - inclusive onward index of current subsequence
                        int integers[], // A - full sequence
                        int nIntegers) 
{
//    printf("value:%4d, hereIndex: %2d, integers[hereIndex]: %3d\n", value, hereIndex, integers[hereIndex]);
    
    // special case: able to express immediately
    if (value == integers[hereIndex]) 
        return 1;
    
    // special case: subsequence has only 1 element
    if (hereIndex == nIntegers - 1)
        return value == integers[hereIndex]; 
    
    // choose the sum with the integer at hereIndex
                        // ____ KEY BELOW HERE!! _____
    if (expressibleBySumOf(value - integers[hereIndex], hereIndex + 1, integers, nIntegers))
        return 1; 
    
    // choose the sum without the integer at hereIndex
    if (expressibleBySumOf(value, hereIndex + 1, integers, nIntegers))
        return 1; 
    
    return 0; 
};

int main(int argc, char** argv) {
    int n, q, i;
    scanf("%d", &n);
    
    const int nIntegers = n;
    int integers[nIntegers];
    
    for (i = 0; i < n; i++)
        scanf("%d", &integers[i]);

//    printf("integers: ");printArray(integers, nIntegers);printf("\n");
    
    scanf("%d", &q);
    
    const int nQuestions = q;
    int questions[nQuestions];
    int answers[nQuestions];
    
    for (i = 0; i < q; i++) {
        scanf("%d", &questions[i]);
//        printf("-------------question: %d--------------\n", questions[i]);
        answers[i] = expressibleBySumOf(questions[i], 0, integers, nIntegers);
//        printf("answer: %d\n",answers[i]);
//        printf("\n");
    }

//    printf("questions: ");printArray(questions, nQuestions);printf("\n");
//    printf("answers:   ");
    printBooleanArray(answers, nQuestions);
    printf("\n");
    

    return (EXIT_SUCCESS);
}