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>
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>*/
 
#define pb push_back
#define mp make_pair
#define sz(s) ((int)(s.size()))
#define all(s) s.begin(), s.end()
#define rep(i, a, n) for (int i = a; i <= n; ++i)
#define per(i, n, a) for (int i = n; i >= a; --i)
#define onlycin ios_base::sync_with_stdio(false); cin.tie(0) 
#define F first
#define S second
using namespace std;
// using namespace __gnu_pbds;
typedef long long ll;
typedef unsigned long long ull;
/*typedef tree<
pair < int, int >,
null_type,
less< pair < int, int > >,
rb_tree_tag,
tree_order_statistics_node_update>
ordered_set;*/
// find_by_order() order_of_key()
const int MAXN = (int)5e5+228;
const char nxtl = '\n';
const int mod = (int)1e9+7;
const double eps = (double)1e-7;
template<typename T> inline bool updmin(T &a, const T &b) {return a > b ? a = b, 1 : 0;}
template<typename T> inline bool updmax(T &a, const T &b) {return a < b ? a = b, 1 : 0;}

int n, a[55], b[55];
bool used[55];
set < int > have;
void dfs(int x) {
	used[x] = 1;
	for(auto &to : have) if(!used[x%to]) dfs(x%to);
}

inline bool can() {
	rep(i, 1, n) {
		memset(used, 0, sizeof used);
		dfs(a[i]);
		if(!used[b[i]]) {
			return 0;
		}
	}
	return 1;
}

int main() {
	#ifdef accepted
		freopen(".in", "r", stdin);
		freopen(".out", "w", stdout);
	#endif
	onlycin;
	cin >> n;
	rep(i, 1, n) cin >> a[i];
	rep(i, 1, n) cin >> b[i];
	ll res = 0;
	rep(i, 1, 50) have.insert(i);
	per(i, 50, 1) {
		have.erase(i);
		if(!can()) {
			res += (1ll << i*1ll);
			have.insert(i);
		}
	}
	if(!can()) res = -1;
	cout << res << nxtl;
	
	return 0;
}