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 Rust (1.15.1) (judge verdict: Accepted)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

fn solve_single(use_set: &BTreeSet<usize>, pow2: &Vec<u64>) -> Vec<Vec<u64>> {
    let max_idx = 51usize;
    let mut dp = vec![vec![std::u64::MAX; max_idx]; max_idx];
    for i in 0..max_idx {
        dp[i][i] = 0;
    }
    for i in 0..max_idx {
        for j in 0..i {
            for &k in use_set.iter() {
                if dp[i % k][j] != std::u64::MAX {
                    let new = dp[i % k][j] + pow2[k];
                    if new < dp[i][j] {
                        dp[i][j] = new;
                    }
                }
            }
        }
    }
    dp
}

fn is_ok(a: &Vec<usize>, b: &Vec<usize>, use_set: &BTreeSet<usize>, pow2: &Vec<u64>) -> bool {
    let dp = solve_single(use_set, pow2);
    (0..a.len()).all(|i| dp[a[i]][b[i]] != std::u64::MAX)
}

fn main() {
    let n = read::<usize>();
    let a = read_vec::<usize>();
    let b = read_vec::<usize>();
    let mut pow2 = vec![0u64; 51];
    pow2[0] = 1;
    for i in 1..51 {
        pow2[i] = pow2[i - 1] * 2;
    }

    let mut use_set = (1..51).collect::<BTreeSet<_>>();
    if !is_ok(&a, &b, &use_set, &pow2) {
        println!("-1");
        return;
    }
    for i in (1..51).rev() {
        use_set.remove(&i);
        if !is_ok(&a, &b, &use_set, &pow2) {
            use_set.insert(i);
        }
    }
    let ans = use_set.into_iter().map(|x| pow2[x]).sum::<u64>();
    println!("{:?}", ans);
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}