Problem:
Score : 700 points

Problem Statement

Aoki is playing with a sequence of numbers a_{1}, a_{2}, ..., a_{N}. Every second, he performs the following operation :

Choose a positive integer k. For each element of the sequence v, Aoki may choose to replace v with its remainder when divided by k, or do nothing with v. The cost of this operation is 2^{k} (regardless of how many elements he changes).

Aoki wants to turn the sequence into b_{1}, b_{2}, ..., b_{N} (the order of the elements is important). Determine if it is possible for Aoki to perform this task and if yes, find the minimum cost required.

Constraints

1 \leq N \leq 50

0 \leq a_{i}, b_{i} \leq 50

All values in the input are integers.

Input

Input is given from Standard Input in the following format:

N
a_{1} a_{2} ... a_{N}
b_{1} b_{2} ... b_{N}

Output

Print the minimum cost required to turn the original sequence into b_{1}, b_{2}, ..., b_{N}. If the task is impossible, output -1 instead.

Sample Input 1

3
19 10 14
0 3 4

Sample Output 1

160

Here's a possible sequence of operations :

Choose k = 7. Replace 19 with 5, 10 with 3 and do nothing to 14. The sequence is now 5, 3, 14.

Choose k = 5. Replace 5 with 0, do nothing to 3 and replace 14 with 4. The sequence is now 0, 3, 4.

The total cost is 2^{7} + 2^{5} = 160.

Sample Input 2

3
19 15 14
0 0 0

Sample Output 2

2

Aoki can just choose k = 1 and turn everything into 0. The cost is 2^{1} = 2.

Sample Input 3

2
8 13
5 13

Sample Output 3

-1

The task is impossible because we can never turn 8 into 5 using the given operation.

Sample Input 4

4
2 0 1 8
2 0 1 8

Sample Output 4

0

Aoki doesn't need to do anything here. The cost is 0.

Sample Input 5

1
50
13

Sample Output 5

137438953472

Beware of overflow issues.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.*;
public class Main {

    static public int n;
    static int[] A = new int[50];
    static int[] B = new int[50];

    public Boolean fun(Long bit){
        boolean[][] arr = new boolean[51][51];
        for(int i = 0; i <= 50; i++) {
                arr[i][i] = true;
        }
        for(int i = 0;i <= 50; i++){
            if((bit & (1<<i)) != 0){
                for(int j = 0;j <= 50;j++){
                    arr[j][j%i] = true;
                }
            }
        }

        for(int k = 0;k <= 50;k++){
            for(int i = 0;i <= 50;i++){
                for(int j = 0;j <= 50;j++){
                    arr[i][j] |= arr[i][k] & arr[k][j];
                }
            }
        }

        Boolean flag = true;

        for(int i = 0;i < n;i++){
            if(!arr[A[i]][B[i]])flag = false;
        }

        return flag;

    }

    public void solve(){
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();

        for(int i = 0;i < n;i++) {
            A[i] = sc.nextInt();
        }
        for(int i = 0;i < n;i++) {
            B[i] = sc.nextInt();
        }

        Long ans = (1L << 50) - 1;

        if(!fun(ans)) {
            System.out.println(-1);
            return;
        }

        for(int i = 50;i >= 0;i--){
            Long tmp = ans & ~(1L << i);
            if(fun(tmp))ans = tmp;
        }

        System.out.println(ans);

    }

    public static void main(String[] args){
        new Main().solve();
    }
};