#line 1 "main.cpp"
#define PROBLEM "https://yukicoder.me/problems/no/1078"
#include <cassert>
#include <iostream>
#line 2 "/home/user/Library/utils/macros.hpp"
#define REP(i, n) for (int i = 0; (i) < (int)(n); ++ (i))
#define REP3(i, m, n) for (int i = (m); (i) < (int)(n); ++ (i))
#define REP_R(i, n) for (int i = (int)(n) - 1; (i) >= 0; -- (i))
#define REP3R(i, m, n) for (int i = (int)(n) - 1; (i) >= (int)(m); -- (i))
#define ALL(x) std::begin(x), std::end(x)
#line 3 "/home/user/Library/utils/two_satisfiability.hpp"
#include <utility>
#include <vector>
#line 2 "/home/user/Library/graph/strongly_connected_components.hpp"
#include <functional>
#line 4 "/home/user/Library/graph/transpose_graph.hpp"

/**
 * @param g is an adjacent list of a digraph
 * @note $O(V + E)$
 * @see https://en.wikipedia.org/wiki/Transpose_graph
 */
std::vector<std::vector<int> > make_transpose_graph(std::vector<std::vector<int> > const & g) {
    int n = g.size();
    std::vector<std::vector<int> > h(n);
    REP (i, n) {
        for (int j : g[i]) {
            h[j].push_back(i);
        }
    }
    return h;
}
#line 7 "/home/user/Library/graph/strongly_connected_components.hpp"

/**
 * @brief strongly connected components decomposition, Kosaraju's algorithm / 強連結成分分解
 * @return the pair (the number k of components, the function from vertices of g to components)
 * @param g is an adjacent list of a digraph
 * @param g_rev is the transpose graph of g
 * @note $O(V + E)$
 */
std::pair<int, std::vector<int> > decompose_to_strongly_connected_components(const std::vector<std::vector<int> > & g, const std::vector<std::vector<int> > & g_rev) {
    int n = g.size();
    std::vector<int> acc(n); {
        std::vector<bool> used(n);
        std::function<void (int)> dfs = [&](int i) {
            used[i] = true;
            for (int j : g[i]) if (not used[j]) dfs(j);
            acc.push_back(i);
        };
        REP (i,n) if (not used[i]) dfs(i);
        reverse(ALL(acc));
    }
    int size = 0;
    std::vector<int> component_of(n); {
        std::vector<bool> used(n);
        std::function<void (int)> rdfs = [&](int i) {
            used[i] = true;
            component_of[i] = size;
            for (int j : g_rev[i]) if (not used[j]) rdfs(j);
        };
        for (int i : acc) if (not used[i]) {
            rdfs(i);
            ++ size;
        }
    }
    return { size, move(component_of) };
}

std::pair<int, std::vector<int> > decompose_to_strongly_connected_components(const std::vector<std::vector<int> > & g) {
    return decompose_to_strongly_connected_components(g, make_transpose_graph(g));
}
#line 6 "/home/user/Library/utils/two_satisfiability.hpp"

/**
 * @brief 2-SAT ($O(N)$)
 * @param n is the number of variables
 * @param cnf is a proposition in a conjunctive normal form. Each literal is expressed as number $x$ s.t. $1 \le \vert x \vert \le n$
 * @return a vector with the length $n$ if SAT. It's empty if UNSAT.
 */
std::vector<bool> compute_two_satisfiability(int n, const std::vector<std::pair<int, int> > & cnf) {
    // make digraph
    std::vector<std::vector<int> > g(2 * n);
    auto index = [&](int x) {
        assert (x != 0 and abs(x) <= n);
        return x > 0 ? x - 1 : n - x - 1;
    };
    for (auto it : cnf) {
        int x, y; std::tie(x, y) = it;  // x or y
        g[index(- x)].push_back(index(y));  // not x implies y
        g[index(- y)].push_back(index(x));  // not y implies x
    }

    // do SCC
    std::vector<int> component = decompose_to_strongly_connected_components(g).second;
    std::vector<bool> valuation(n);
    REP3 (x, 1, n + 1) {
        if (component[index(x)] == component[index(- x)]) {  // x iff not x
            return std::vector<bool>();  // unsat
        }
        valuation[x - 1] = component[index(x)] > component[index(- x)];  // use components which indices are large
    }
    return valuation;
}
#line 6 "main.cpp"
using namespace std;

vector<vector<bool> > solve(int n, vector<int> & s, vector<int> & t, vector<int> & u) {
    auto var = [&](int y, int x) { return y * n + x; };
    vector<pair<int, int> > cnf;
    REP (y, n) {
        REP (x, n) {
            int a = var(s[y], x) + 1;
            int b = var(x, t[y]) + 1;
            if (u[y] == 0) {
                cnf.emplace_back(a, b);
            } else if (u[y] == 1) {
                cnf.emplace_back(- a, b);
            } else if (u[y] == 2) {
                cnf.emplace_back(a, - b);
            } else if (u[y] == 3) {
                cnf.emplace_back(- a, - b);
            } else {
                assert (false);
            }
        }
    }
    auto ev = compute_two_satisfiability(n * n, cnf);
    if (ev.empty()) {
        return vector<vector<bool> >();
    } else {
        vector<vector<bool> > a(n, vector<bool>(n));
        REP (y, n) {
            REP (x, n) {
                a[y][x] = ev[var(y, x)];
            }
        }
        return a;
    }
}

int main() {
    // input
    int n; scanf("%d", &n);
    vector<int> s(n);
    REP (i, n) {
        scanf("%d", &s[i]);
        -- s[i];
    }
    vector<int> t(n);
    REP (i, n) {
        scanf("%d", &t[i]);
        -- t[i];
    }
    vector<int> u(n);
    REP (i, n) {
        scanf("%d", &u[i]);
    }

    // solve
    auto a = solve(n, s, t, u);

    // output
    if (a.empty()) {
        printf("%d\n", -1);
    } else {
        REP (y, n) {
            REP (x, n) {
                printf("%d%c", (int)a[y][x], x + 1 < n ? ' ' : '\n');
            }
        }
    }
    return 0;
}