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 <vector>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
using ll = long long;

int main() {
    int n;
    cin >> n;

    vector<int> a(n), b(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    for (int i = 0; i < n; i++) {
        cin >> b[i];
    }

    constexpr int K = 50;
    ll m = (1LL << (K + 1)) - 1;

    for (int l = K; l >= 0; l--) {
        m ^= 1LL << l;

        int dp[K + 1][K + 1] = {}; // iをjに変えられるなら1
        for (int i = 0; i <= K; i++) {
            dp[i][i] = 1;
        }
        for (int k = 1; k < K; k++) {
            if (m & 1LL << k) {
                for (int i = 0; i <= K; i++) {
                    dp[i][i % k] = 1;
                }
            }
        }
        for (int k = 1; k < K; k++) {
            if (m & 1LL << k) {
                for (int i = 0; i <= K; i++) {
                    for (int j = 0; j < i; j++) {
                        dp[i][j] |= dp[i][i % k] & dp[i % k][j];
                    }
                }
            }
        }

        int t = 1;
        for (int i = 0; i < n; i++) {
            if (dp[a[i]][b[i]] == 0) {
                t = 0;
                break;
            }
        }
        if (t == 0) {
            m ^= 1LL << l;
            if (l == K) break;
        }
    }

    if (m >= 1LL << K) {
        cout << -1 << endl;
    } else {
        cout << m << endl;
    }

    return 0;
}