#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>

using namespace std;

// Helper function to compute the Greatest Common Divisor
long long gcd(long long a, long long b) {
    while (b) {
        a %= b;
        swap(a, b);
    }
    return a;
}

int main() {
    // Optimize standard I/O operations for performance
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int n;
    if (!(cin >> n)) return 0;

    vector<long long> a(n + 1), b(n + 1);
    long long g = 0;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        g = gcd(g, a[i]);
    }
    for (int i = 1; i <= n; ++i) {
        cin >> b[i];
    }

    // If the overall GCD is already 1, no operations are required.
    if (g == 1) {
        cout << 0 << "\n";
        return 0;
    }

    // Extract unique prime factors of the initial overall GCD
    vector<long long> primes;
    long long temp = g;
    for (long long i = 2; i * i <= temp; ++i) {
        if (temp % i == 0) {
            primes.push_back(i);
            while (temp % i == 0) {
                temp /= i;
            }
        }
    }
    if (temp > 1) {
        primes.push_back(temp);
    }

    int k = primes.size();
    int max_mask = 1 << k;
    
    // dp[mask] stores the minimum cost to achieve the state represented by mask
    // -1 denotes an unreachable state
    vector<long long> dp(max_mask, -1);
    dp[0] = 0;

    for (int i = 1; i <= n; ++i) {
        int mask_i = 0;
        for (int j = 0; j < k; ++j) {
            if (b[i] % primes[j] != 0) {
                mask_i |= (1 << j);
            }
        }
        
        // If this element eliminates no prime factors, skip it
        if (mask_i == 0) continue;

        long long cost = 1LL * i * i + i;
        
        // Update the DP table from top to bottom to use only one array copy
        for (int mask = max_mask - 1; mask >= 0; --mask) {
            if (dp[mask] != -1) {
                int next_mask = mask | mask_i;
                if (dp[next_mask] == -1 || dp[mask] + cost < dp[next_mask]) {
                    dp[next_mask] = dp[mask] + cost;
                }
            }
        }
    }

    // Output the minimum cost to eliminate all prime factors, or -1 if impossible
    cout << dp[max_mask - 1] << "\n";

    return 0;
}