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 <iostream>
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
ll graph[55][55][55];

int main()
{
    ll N;
    cin>>N;
    ll a[N],b[N];
    vector<ll> c[N];
    for(ll i=0; i<N; i++){
        cin>>a[i];
        c[i].push_back(a[i]);
    }
    for(ll i=0; i<N; i++){
        cin>>b[i];
    }

    for(ll i=0; i<55; i++){
        for(ll j=0; j<55; j++){
            graph[i][i][j] = 1;
        }
    }


    for(ll k=1; k<55; k++){
        for(ll i=0; i<55; i++){
            for(ll j=0; j<55; j++){
                if(graph[i%k][j][k-1]==1 || graph[i][j][k-1]==1)
                    graph[i][j][k] = 1;
            }
        }
    }


    for(ll i=0; i<N; i++){
        if(graph[a[i]][b[i]][50]==0){
            cout<<-1<<endl;
            return 0;
        }
    }

    ll ans = 0;
    for(ll k=50; k>0; k--){
        bool check = true;
        for(ll i=0; i<N; i++){
            bool test = false;
            for(ll j=0; j<c[i].size(); j++){
                if(graph[c[i][j]][b[i]][k-1]==1){
                    test = true;
                }
            }
            if(!test)
                check = false;

        }
        if(!check){
            ans+= (1ll<<k);
            for(ll i=0; i<N; i++){
                vector<ll> carry;
                ll poss[55];
                for(ll j=0; j<55; j++)
                    poss[j] = 0;
                for(ll j=0; j<c[i].size(); j++){
                    poss[c[i][j]] = 1;
                    poss[c[i][j]%k] = 1;
                }
                for(ll j=0; j<55; j++){
                    if(poss[j]==1)
                        carry.push_back(j);
                }
                c[i] = carry;
            }

        }

    }
    cout<<ans<<endl;

    return 0;
}