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 Go (judge verdict: Accepted)

package main

import (
	"bufio"
	"bytes"
	"fmt"
	"os"
	"strconv"
	"strings"
)

var (
	s = bufio.NewScanner(os.Stdin)
	w = bytes.Buffer{}
	n = 0
	a = []int{}
)

func main() {
	n = nextInt(s)
	a = atoi(s)
	_ = nextInt(s)
	m := atoi(s)

	for _, v := range m {
		i := 0
		if solve(i, v) {
			w.WriteString("yes\n")
			continue
		}
		w.WriteString("no\n")
	}
	fmt.Println(strings.TrimRight(w.String(), "\n"))
}

func nextInt(s *bufio.Scanner) int {
	s.Scan()
	r, err := strconv.Atoi(s.Text())
	if err != nil {
		panic(err)
	}
	return r
}

func atoi(s *bufio.Scanner) []int {
	s.Scan()
	a := strings.Fields(s.Text())
	i := []int{}

	for _, v := range a {
		j, err := strconv.Atoi(v)
		if err != nil {
			panic(err)
		}
		i = append(i, j)
	}
	return i
}

func solve(i, m int) bool {
	if m == 0 {
		return true
	} else if i >= n {
		return false
	}
	return solve(i+1, m) || solve(i+1, m-a[i])
}