Problem:
Score : 600 points

Problem Statement

Given are an integer N and arrays S, T, U, and V, each of length N.
Construct an N×N matrix a that satisfy the following conditions:

a_{i,j} is an integer.

0 \leq a_{i,j} \lt 2^{64}.

If S_{i} = 0, the bitwise AND of the elements in the i-th row is U_{i}.

If S_{i} = 1, the bitwise OR of the elements in the i-th row is U_{i}.

If T_{i} = 0, the bitwise AND of the elements in the i-th column is V_{i}.

If T_{i} = 1, the bitwise OR of the elements in the i-th column is V_{i}.

However, there may be cases where no matrix satisfies the conditions.

Constraints

All values in input are integers.

1 \leq N \leq 500

0 \leq S_{i} \leq 1

0 \leq T_{i} \leq 1

0 \leq U_{i} \lt 2^{64}

0 \leq V_{i} \lt 2^{64}

Input

Input is given from Standard Input in the following format:

N
S_{1} S_{2} ... S_{N}
T_{1} T_{2} ... T_{N}
U_{1} U_{2} ... U_{N}
V_{1} V_{2} ... V_{N}

Output

If there exists a matrix that satisfies the conditions, print one such matrix in the following format:

a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Note that any matrix satisfying the conditions is accepted.

If no matrix satisfies the conditions, print -1.

Sample Input 1

2
0 1
1 0
1 1
1 0

Sample Output 1

1 1
1 0

In Sample Input 1, we need to find a matrix such that:

the bitwise AND of the elements in the 1-st row is 1;

the bitwise OR of the elements in the 2-nd row is 1;

the bitwise OR of the elements in the 1-st column is 1;

the bitwise AND of the elements in the 2-nd column is 0.

Sample Input 2

2
1 1
1 0
15 15
15 11

Sample Output 2

15 11
15 11

Solution in PyPy3 (7.3.0) (judge verdict: Runtime Error)

import heapq


def inpl():
    return list(map(int, input().split()))


N, M, S = inpl()
A = 50  # max A
S = min(S, M * A)

lines = [[] for _ in range(N * (M * A + 1))]
for _ in range(M):
    in_s, in_t, cost, time = inpl()
    in_s -= 1
    in_t -= 1

    for n in range(cost, M * A + 1):
        s = in_s + n * N
        t = in_t + (n - cost) * N
        lines[s].append((t, time))

        s = in_s + (n - cost) * N
        t = in_t + n * N
        lines[t].append((s, time))


for i in range(N):
    cost, time = inpl()
    for n in range(0, M * A + 1):
        s = i + n * N
        if n + cost > M * A:
            t = i + (M * A) * N
        else:
            t = i + (n + cost) * N
        lines[s].append((t, time))


INF = 10**20


def dijkstra(N, S):
    weight = [INF] * N
    weight[S] = 0
    q = [[0, S]]
    heapq.heapify(q)
    while q:
        w0, s = heapq.heappop(q)
        for t, w in lines[s]:
            w += w0
            if weight[t] > w:
                heapq.heappush(q, [w, t])
                weight[t] = w

    return weight


weight = dijkstra(N * (M * A + 1), 0 + S * N)

print("\n".join(map(str, weight[1:N])))