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 Python3 (judge verdict: Wrong Answer)

# -*- coding: utf-8 -*-

import random
import sys
import os
import pprint

#fd = os.open('ALDS1_5_A.txt', os.O_RDONLY)
#os.dup2(fd, sys.stdin.fileno())

n = int(input())
A = list(map(int, input().split()))

# memo table T[i][m]
row_num = len(A)
column_num = 2000
T = [[None for i in range(column_num)] for j in range(row_num)]
#pprint.pprint(T)


def solve(i, m):
    # out of index
    if i >= len(A):
        return False
    if m < 0:
        return False

    if T[i][m] is not None:
        return T[i][m]

    if m == 0:
        T[i][m] = True
    #elif i >= len(A):
    #    T[i][m] = False
    elif solve(i+1, m):
        T[i][m] = True
    elif solve(i+1, m - A[i]):
        T[i][m] = True
    else:
        T[i][m] = False
    return T[i][m]

test_num = int(input())
test_values = map(int, input().split())
for v in test_values:
    result = solve(0, v)
    if result is True:
        print('yes')
    else:
        print('no')