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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define ll long long
using namespace std;

int n,a[55],b[55];
bool v[55][55];
ll ans=0;

inline bool can(){
	memset(v,0,sizeof(v));
	for(int i=1;i<=50;i++) if((1ll<<i)&ans)
		for(int j=0;j<=50;j++) v[j][j%i]=1;

	for(int i=0;i<=50;i++) v[i][i]=1;

	for(int k=0;k<=50;k++)
		for(int i=0;i<=50;i++)
		    for(int j=0;j<=50;j++) if(v[i][k]&&v[k][j]) v[i][j]=1;

	for(int i=0;i<n;i++) if(!v[a[i]][b[i]]) return 0;
	return 1;
}

int main(){
	scanf("%d",&n),ans=1ll<<51,ans-=2;
	for(int i=0;i<n;i++) scanf("%d",a+i);
	for(int i=0;i<n;i++) scanf("%d",b+i);

	if(!can()){ puts("-1"); return 0;}

	for(int i=50;i;i--){
		ans^=1ll<<i;
		if(!can()) ans|=1ll<<i;
	}

	printf("%lld\n",ans);
	return 0;
}