Problem: 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.

Examples

Input

2
0 1
1 0
1 1
1 0


Output

1 1
1 0


Input

2
1 1
1 0
15 15
15 11


Output

15 11
15 11

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using ull = unsigned long long;
using P = pair<int,int>;

vector<int> s[2];
vector<ull> u[2];

const int N = 500;
int n;
int val[2][N];
int d[N][N];

void flip() {
    rep(i,n)rep(j,i) {
        swap(d[i][j], d[j][i]);
    }
}
bool solve() {
    rep(i,n)rep(j,n) d[i][j] = -1;
    rep(k,2) {
        rep(i,n) {
            int x = val[k][i];
            if (s[k][i] != x) {//0 and    1   or
                rep(j,n) {
                    if (d[i][j] == !x) return false;
                    d[i][j] = x;
                }
            }
        }
        flip();
    }

    bool changed = false;
    /*rep(_,2) */{
        rep(k,2) {
            rep(i,n) {
                int x = val[k][i];
                if (s[k][i] == x) {//0 and    1   or
                    vector<int> p;
                    bool ok = false;
                    rep(j,n) {
                        if (d[i][j] == -1) p.push_back(j);
                        if (d[i][j] == x) ok = true;
                    }
                    if (ok) continue;
                    if (p.size() == 0) return false;
                    if (p.size() == 1) {
                        d[i][p[0]] = x;
                        changed = true;
                    }
                }
            }
            flip();
        }
    }

    if (changed){
        rep(k,2) {
            if (k==0)
            rep(i,n) {
                int x = val[k][i];
                if (s[k][i] == x) {//0 and    1   or
                    vector<int> p;
                    bool ok = false;
                    rep(j,n) {
                        if (d[i][j] == -1) p.push_back(j);
                        if (d[i][j] == x) ok = true;
                    }
                    if (ok) continue;
                    if (p.size() == 0) return false;//need this the second iteration
                    if (p.size() == 1) {
                        d[i][p[0]] = x;
                    }
                }
            }
            flip();
        }
    }

    vector<int> is, js;
    rep(i,n) {
        bool filled = true;
        rep(j,n) if (d[i][j] == -1) filled = false;
        if (!filled) is.push_back(i);
    }
    rep(j,n) {
        bool filled = true;
        rep(i,n) if (d[i][j] == -1) filled = false;
        if (!filled) js.push_back(j);
    }
    rep(i,is.size())rep(j,js.size()) {
        d[is[i]][js[j]] = (i+j)%2;
    }
    return true;
}

ull ans[N][N];

int main() {
    cin >> n;
    rep(i,2) {
        s[i] = vector<int>(n);
        rep(j,n) cin >> s[i][j];
    }
    rep(i,2) {
        u[i] = vector<ull>(n);
        rep(j,n) cin >> u[i][j];
    }

    rep(b,64) {
        rep(i,2)rep(j,n) val[i][j] = u[i][j]>>b&1;
        if (!solve()) {
            cout << "-1" << endl;
            return 0;
        }
        rep(i,n)rep(j,n) ans[i][j] |= ull(d[i][j])<<b;
    }
    rep(i,n) {
        rep(j,n) {
            cout<<ans[i][j]<<(j==n-1?'\n':' ');
        }
    }
    return 0;
}