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>
using namespace std;
using ull = uint64_t;

int n;
bool s[500], t[500], u[500], v[500];
int a[500][500];
ull ans[500][500];
void end(){
    cout << -1 << endl;
    exit(0);
}
void solve(){
    memset(a, 0xff, sizeof(a));
    list<int> x(n), y(n);
    auto px = x.begin(), py = y.begin();
    iota(x.begin(), x.end(), 0);
    iota(y.begin(), y.end(), 0);
    for(int i = 0; i < n; i++){
        if(s[i] ^ u[i]){
            for(int j = 0; j < n; j++) a[i][j] = u[i];
            px = x.erase(px);
        }
        else px++;
    }
    for(int j = 0; j < n; j++){
        if(t[j] ^ v[j]){
            for(int i = 0; i < n; i++){
                if(a[i][j] == (v[j] ^ 1)) end();
                a[i][j] = v[j];
            }
            py = y.erase(py);
        }
        else py++;
    }
    if(x.size() > 1 && y.size() > 1){
        bool b = 0;
        for(int i : x){
            bool c = b;
            for(int j : y){
                a[i][j] = c;
                c = !c;
            }
            b = !b;
        }
    }
    else if(x.size() == 1){
        const int i = x.front();
        for(int j : y){
            a[i][j] = u[i];
            bool flag = 1;
            for(int i = 0; i < n; i++) if(a[i][j] == v[j]) flag = 0;
            if(flag) a[i][j] = v[j];
        }
    }
    else if(y.size() == 1){
        const int j = y.front();
        for(int i : x){
            a[i][j] = v[j];
            bool flag = 1;
            for(int j = 0; j < n; j++) if(a[i][j] == u[i]) flag = 0;
            if(flag) a[i][j] = u[i];
        }
    }
    for(int i : x){
        bool flag = 1;
        for(int j = 0; j < n; j++) if(a[i][j] == u[i]) flag = 0;
        if(flag) end();
    }
    for(int j : y){
        bool flag = 1;
        for(int i = 0; i < n; i++) if(a[i][j] == v[j]) flag = 0;
        if(flag) end();
    }
}
int main(){
    cin >> n;
    vector<ull> u_(n), v_(n);
    for(int i = 0; i < n; i++) cin >> s[i];
    for(int i = 0; i < n; i++) cin >> t[i];
    for(auto& i : u_) cin >> i;
    for(auto& i : v_) cin >> i;
    for(int bit = 64; bit--; ){
        for(int i = 0; i < n; i++) u[i] = u_[i] >> bit & 1;
        for(int i = 0; i < n; i++) v[i] = v_[i] >> bit & 1;
        solve();
        for(int i = 0; i < n; i++) for(int j = 0; j < n; j++){
            ans[i][j] <<= 1;
            ans[i][j] |= a[i][j];
        }
    }
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++) cout << ans[i][j] << ' ';
        cout << '\n';
    }
}