#include <bits/stdc++.h>
#define ms(x, v) memset(x, v, sizeof(x))
#define il __attribute__((always_inline)) static
#define U(i,l,r) for(int i(l),END##i(r);i<=END##i;++i)
#define D(i,r,l) for(int i(r),END##i(l);i>=END##i;--i)
using namespace std;

typedef unsigned long long ull;
typedef long long ll;
template <class T> using BS = basic_string<T>;

template <class T> void rd(T& s) {
	int c = getchar(); T f = 1; s = 0;
	while (!isdigit(c)) { if (c == '-') f = -1; c = getchar(); }
	while (isdigit(c)) { s = s * 10 + (c ^ 48); c = getchar(); }
	s *= f;
}
template <class T, class... Y> void rd(T& x, Y&... y) { rd(x), rd(y...); }
#define meow(...) fprintf(stderr, __VA_ARGS__)
#define Assert(e, v) if (!(e)) exit(v);

const int N = 300005;
int n, p[N], q[N], a[N];
bool vis[N], beg[N], le[N]; int shift[N];
vector<pair<int, int>> ri[N];

// 只需检查两个同余方程组是否相容
vector<pair<int, int>> equ; // x \equiv xi (mi)
// 加入一个新长度的环，一定存在 xi 使得有解

bool check(int x1, int m1, int x2, int m2) {
    int g = __gcd(m1, m2);
    return !(((x2 - x1) % g + g) % g);
}
bool tryMerge(int x1, int m1) {
    for (auto pi : equ) {
        int x2, m2; tie(x2, m2) = pi;
        if (!check(x1, m1, x2, m2))
            return 0;
    }
    return 1;
}
int main() {
    rd(n);
    U (i, 1, n) rd(p[i]);
    U (i, 1, n) { rd(q[i]); a[q[i]] = i; }

    U (i, 1, n) if (!vis[i]) {
        beg[i] = 1;
        int t = i, s = 0;
        while (!vis[t]) {
            vis[t] = 1;
            ri[i].push_back({p[t], s});
            t = q[t]; ++s;
        }
    }
    // 最多有根号个不同的环长. 确定一个该长度的环，其余该长度的环已经确定
    U (i, 1, n) sort(ri[i].begin(), ri[i].end());

    U (i, 1, n) if (beg[i] && !le[ri[i].size()]) {
        int m = ri[i].size();
        le[m] = 1;
        for (auto pi : ri[i]) {
            int v, s; tie(v, s) = pi;
            if (tryMerge(s, m)) {
                equ.emplace_back(s, m);
                shift[m] = s;
                break;
            }
        }
    }
    int ans[N] {};
    U (i, 1, n)
        sort(ri[i].begin(), ri[i].end(), [](const pair<int, int> &x, const pair<int, int> &y) {
            return x.second < y.second;
        });
    U (i, 1, n) if (beg[i]) {
        int m = ri[i].size();
        int t = i;
        U (j, shift[m], m - 1) {
            ans[t] = ri[i][j].first;
            t = q[t];
        }
        U (j, 0, shift[m] - 1) {
            ans[t] = ri[i][j].first;
            t = q[t];
        }
    }
    U (i, 1, n) printf("%d ", ans[i]);
    puts("");
}