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 (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

int main()
{
	int i, N, a[51], b[51];
	scanf("%d", &N);
	for (i = 1; i <= N; i++) scanf("%d", &(a[i]));
	for (i = 1; i <= N; i++) scanf("%d", &(b[i]));
	for (i = 1; i <= N; i++) if (b[i] != a[i]) break;
	if (i > N) {
		printf("0\n");
		fflush(stdout);
		return 0;
	}
	for (i = 1; i <= N; i++) if (b[i] != a[i] && b[i] * 2 >= a[i]) break;
	if (i <= N) {
		printf("-1\n");
		fflush(stdout);
		return 0;
	}
	
	int j, k, l, adj[51][51] = {};
	long long bit[51], ans;
	for (i = 1, bit[0] = 1; i <= 50; i++) bit[i] = bit[i-1] << 1;
	for (i = 49, ans = bit[50] - 2; i >= 1; i--) {
		ans ^= bit[i];
		for (j = 0; j <= 50; j++) for (k = 0, adj[j][j] = 1; k < j; k++) adj[j][k] = 0;
		for (j = 1; j < 50; j++) {
			if ((ans & bit[j]) == 0) continue;
			for (k = j; k <= 50; k++) adj[k][k%j] = 1;
		}
		for (l = 1; l <= 50; l++) {
			for (j = l; j <= 50; j++) {
				for (k = 0; k <= l; k++) if ((adj[j][l] & adj[l][k]) == 1) adj[j][k] = 1;
			}
		}
		for (j = 1; j <= N; j++) if (adj[a[j]][b[j]] == 0) break;
		if (j <= N) ans ^= bit[i];
	}
	
	printf("%lld\n", ans);
	fflush(stdout);
	return 0;
}