For this Question: You are given two arrays A and B, both of size N. Let S = { S<sub>1</sub>,S<sub>2</sub>,S<sub>3</sub>,..., S<sub>m</sub> } be any subset of integers from the set { 1, 2, 3,..., N }. We define:

L(S) = lcm(A<sub>S<sub>1</sub></sub>, A<sub>S<sub>2</sub></sub>, ..., A<sub>S<sub>m</sub></sub>) 
V(S) = B<sub>S<sub>1</sub></sub> + B<sub>S<sub>2</sub></sub> + ... + B<sub>S<sub>m</sub></sub>

Finally, we define f(k) as the minimum possible value of V(S) over all subsets S such that L(S) = k; if no such S exists then f(k) is defined to be 0. 

Return the bitwise-xor of f(k) for all k from 1 to 10<sup>6</sup>.The first line of the input consists of a single integer N (1 &le;  N &le;  10<sup>6</sup>).
The second line contains N space-separated integers A<sub>1</sub>, A<sub>2</sub>, ... A<sub>N</sub> (1 &le;  A<sub>i</sub> &le;  10<sup>6</sup>).
The third line contains N space-separated integers B<sub>1</sub>, B<sub>2</sub>, ... B<sub>N</sub> (1 &le;  B<sub>i</sub> &le;  10<sup>6</sup>).Print a single integer denoting the required answer.Sample Input 1:
2
6 1
2 3

Sample Output 1:
1

Sample Explanation 1:
The value of f(1) = 3, f(6) = 2. For all other k, f(k) =0, so answer = 3&oplus;2 = 1.


Sample Input 2:
3
2 8 6
92 15 80

Sample Output 2:
92, I have written this Solution Code: #include <bits/stdc++.h>
#define endl '\n'
using namespace std;
 
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
 
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define aint(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const int inf = 1e9;
const int mod = 1e9 + 7;
//const int mod = 998244353;
const int N = 1e6;
 
vi lpf(N + 1), lpfull(N + 1); 
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    
    read(n);
    readarr(a, n);
    readarr(b, n);
    vi best(N + 1, inf);
    FOR (i, 1, n) best[a[i]] = min(best[a[i]], b[i]);
 
    vvi dp(N + 1, vi(0));
    FOR (i, 2, N)
    {
        if (!lpf[i])
            for (int j = i; j <= N; j += i) lpf[j] = i;
 
        int x = lpf[i];
        if (lpf[i/x] == x) lpfull[i] = lpfull[i/x]*x;
        else lpfull[i] = x;
    }
    
    int first_zero_bit[1001] = {};
    FOR (i, 1, 1000)
    {
        FOR (j, 0, 30)
            if (!((i >> j) & 1)) {first_zero_bit[i] = j; break;}
    }
 
    dp[1].pb(inf);
    int ans = 0;
    if (best[1] < inf) ans = best[1];
 
    FOR (i, 2, N)
    {
        vi v;
        int temp = i;
        while (temp > 1)
        {
            v.pb(lpf[temp]);
            temp /= lpfull[temp];
        }
        temp = 1 << sz(v);
 
        dp[i].assign(temp, inf);
        FORD (j, temp - 2, 0)
        {
            int augment = (1 << first_zero_bit[j]) + j;
            int new_i = i/v[first_zero_bit[j]], new_j = j;
            if (new_i % v[first_zero_bit[j]])
            { 
                if (first_zero_bit[j]) new_j = j - (j >> first_zero_bit[j])*(1 << (first_zero_bit[j] - 1));
                else new_j = j/2;
            }
 
            dp[i][j] = min({dp[i][augment], dp[new_i][new_j], best[new_i]});
        }
 
        FOR (j, 0, temp/2 - 1)
            best[i] = min(best[i], dp[i][j] + dp[i][temp - j - 1]);
 
        if (best[i] < inf) ans ^= best[i];
    }
 
    print(ans);
}
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
, In this Programming Language: C++, Now tell me if this Code is compilable or not?