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 <string>
#include <cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#define _USE_MATH_DEFINES
#include <math.h>
#include <functional>
using namespace std;

#define rep(i,x) for(ll i=0;i<x;i++)
#define repn(i,x) for(ll i=1;i<=x;i++)

typedef long long ll;
const ll INF = 1e17;
const ll MOD = 1000000007;
const ll MAX = 4000001;

ll max(ll a, ll b) {
	if (a > b) { return a; }
	return b;
}

ll min(ll a, ll b) {
	if (a > b) { return b; }
	return a;
}

ll gcd(ll a, ll b) {
	if (b == 0) { return a; }
	if (a < b) { return gcd(b, a); }
	return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}

struct edge {
	ll ind;
	ll fr;
	ll to;
	ll d;
};

class mint {
    long long x;
public:
    mint(long long x = 0) : x((x% MOD + MOD) % MOD) {}
    mint operator-() const {
        return mint(-x);
    }
    mint& operator+=(const mint& a) {
        if ((x += a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator-=(const mint& a) {
        if ((x += MOD - a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator*=(const  mint& a) {
        (x *= a.x) %= MOD;
        return *this;
    }
    mint operator+(const mint& a) const {
        mint res(*this);
        return res += a;
    }
    mint operator-(const mint& a) const {
        mint res(*this);
        return res -= a;
    }
    mint operator*(const mint& a) const {
        mint res(*this);
        return res *= a;
    }

    mint operator^(const ll& t)const {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }

    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }
    // for prime MOD
    mint inv() const {
        return pow(MOD - 2);
    }
    mint& operator/=(const mint& a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint& a) const {
        mint res(*this);
        return res /= a;
    }

    friend ostream& operator<<(ostream& os, const mint& m) {
        os << m.x;
        return os;
    }
};

//////////////////////////


int main() {
    ll N;
    cin >> N;
    
    vector<ll> a(N + 1), b(N + 1);
    repn(i, N) cin >> a[i];
    repn(i, N) cin >> b[i];

    vector<ll> sc(51);
    
    for (ll i = 50; i >= 0; i--) {
        vector<ll> d(51,0);
        repn(j, i - 1) { d[j] = 1; }
        for (ll j = i + 1; j <= 50; j++) { d[j] = sc[j]; }

        vector<vector<bool>> dp(51, vector<bool>(51, 0));

        rep(j, 51) {
            dp[j][j] = 1;

            for (ll k = j; k >= 0; k--) {
                for (ll l = k + 1; l < j; l++) {
                    if (dp[j][l] && dp[l][k]) { dp[j][k] = 1; }
                }
                repn(l, 50) {
                    if (d[l] == 1 && j % l == k) { dp[j][k] = 1; }
                }

            }
        }

        bool bo = 1;
        repn(j, N) {
            if (!dp[a[j]][b[j]]) { bo = 0; }
        }

        if (bo == 1) { sc[i] = 0; }
        else { sc[i] = 1; }

       //cout << i << sc[i] << endl;

    }

    ll ans = 0;
    repn(i, 50) {
        ans += ((ll)1 << i) * sc[i];
    }

    if (sc[0] == 1) { ans = -1; }
    cout << ans;

	system("PAUSE");
}