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 rep(i,n) for(int i=0; i<(n); i++)
#define reps(i,x,n) for(int i=x; i<(n); i++)
#define rrep(i,n) for(int i=(n)-1; i>=0; i--)
#define all(X) (X).begin(),(X).end()
#define X first
#define Y second
#define pb push_back
#define eb emplace_back

using namespace std;
typedef long long int ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

template<class A, size_t N, class T> void Fill(A (&a)[N], const T &v){ fill( (T*)a, (T*)(a+N), v ); }

template<class T> ostream& operator<<(ostream &os, const vector<T> &t) {os<<"{"; rep(i,t.size()) {os<<t[i]<<",";} os<<"}"; return os;}
template<class T,size_t n> ostream& operator<<(ostream &os, const array<T,n> &t) {os<<"{"; rep(i,n) {os<<t[i]<<",";} os<<"}"; return os;}
template<class S, class T> ostream& operator<<(ostream &os, const pair<S,T> &t) { return os<<"("<<t.first<<","<<t.second<<")";}

const ll INF = 1e9+7;


int main(){
	ios_base::sync_with_stdio(false);
	ll N, ans=1e17;
	ll a[55], b[55];

	cin >> N;
	rep(i,N) cin >> a[i];
	rep(i,N) cin >> b[i];

	rep(i,N) if( a[i] != b[i] && a[i] <= b[i]*2 ){
		cout << -1 << endl;
		return 0;
	}

	ll cost = (1ll<<50)-1;
	rrep(n,52) if( cost & (1ll<<n) ){
		ll k = cost ^ (1ll<<n);
		vector<int> v;
		rep(i,52) if( k&(1ll<<i) ){
			v.push_back(i+1);
		}
		ll dp[52]={};
		rep(i,51) dp[i] = (1ll<<i);
		reps(i,1,51) for(auto t: v){
			dp[i] |= dp[i%t];
		}
		bool ok = true;
		rep(i,N) if( !(dp[a[i]] & (1ll << b[i])) ){
			ok = false;
		}
		if(ok) cost = k;
	}
	ans = cost * 2ll;

	cout << ans << endl;

	return 0;
}