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<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long ll;
const int maxn = 50 + 2;

int n, a[maxn], b[maxn], used[maxn];
ll Ans;

bool dp[maxn][maxn];
inline bool Check(int id) {
  memset(dp,0,sizeof(dp));
	dp[50][a[id]] = 1;
	for(int i = 50;i >= 1;i --) {
		for(int j = 0;j <= a[id];j ++) {
			if(!dp[i][j]) continue;
			dp[i - 1][j] = 1;
			if(used[i]) dp[i - 1][j % i] = 1;
		}
	}
	return dp[0][b[id]];
}

inline bool check(int x) {
	int flag = 1;
	used[x] = 0;
	if(x == 1) {
		int xxx = 1;
		xxx ++;
	}
	for(int i = 1;i <= n;i ++) {
		flag &= Check(i);
	}
	used[x] = 1;
	return flag;
}

int main() {
	scanf("%d", &n);
	for(int i = 1;i <= n;i ++) scanf("%d", &a[i]);
	for(int i = 1;i <= n;i ++) scanf("%d", &b[i]);
	for(int i = 1;i <= n;i ++) {
		if(a[i] != b[i] && b[i] * 2 >= a[i]) {
			puts("-1");
			return 0;
		}
	}
	Ans = 0;
	for(int i = 0;i <= 50;i ++) used[i] = 1;
	for(int i = 50;i >= 0;i --) {
    if(!check(i)) Ans += 1LL << i;
		else used[i] = 0;
	}
	printf("%lld\n", Ans);
	return 0;
}