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 ll unsigned long long
using namespace std;
const int N = 505;
int S[N], T[N], n;
ll U[N], V[N], ans[N][N];
int t[N][N], R[N][2], C[N][2];
bool solve() {
    for(int k = 0; k < 64; k++) {
        memset(t, -1, sizeof t);
        memset(R, 0, sizeof R);
        memset(C, 0, sizeof C);
        for(int i = 1; i <= n; i++) {
            int x = (U[i] & 1);
            if(S[i] == x) continue;
            for(int j = 1; j <= n; j++) {
                if(~t[i][j] && t[i][j] != x) return false;
                t[i][j] = x;
            }
        }
        for(int i = 1; i <= n; i++) {
            int x = (V[i] & 1);
            if(T[i] == x) continue;
            for(int j = 1; j <= n; j++) {
                if(~t[j][i] && t[j][i] != x) return false;
                t[j][i] = x;
            }
        }
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                if((U[i] & 1) == (V[j] & 1)) t[i][j] = (U[i] & 1);
                if(t[i][j] == -1) t[i][j] = 0;
                R[i][t[i][j]]++;
                C[j][t[i][j]]++;
            }
        }
        for(int i = 1; i <= n; i++) {
            if(S[i] && (U[i] & 1)) {// | 1
                if(R[i][1]) continue;
                for(int j = 1; j <= n; j++) {
                    if(!T[j] && !(V[j] & 1) && C[j][0] > 1) {//& 0 已有0 
                        t[i][j] = 1;
                        R[i][0]--, C[j][0]--;
                        R[i][1]++, C[j][1]++;
                        break;
                    }
                }
            }
        }
        for(int i = 1; i <= n; i++) {
            if(T[i] && (V[i] & 1)) {
                if(C[i][1]) continue;
                for(int j = 1; j <= n; j++) {
                    if(!S[j] && !(U[j] & 1) && R[j][0] > 1) {
                        t[j][i] = 1;
                        R[j][0]--, C[i][0]--;
                        R[j][1]++, C[i][1]++;
                        break;
                    }
                }
            }
        }
        for(int i = 1; i <= n; i++) {
            if(S[i] == 0 && (U[i] & 1) == 1 && R[i][1] != n) return false;
            if(S[i] == 0 && (U[i] & 1) == 0 && R[i][1] == n) return false;
            if(S[i] == 1 && (U[i] & 1) == 1 && R[i][1] == 0) return false;
            if(S[i] == 1 && (U[i] & 1) == 0 && R[i][1] != 0) return false;
            if(T[i] == 0 && (V[i] & 1) == 1 && C[i][1] != n) return false;
            if(T[i] == 0 && (V[i] & 1) == 0 && C[i][1] == n) return false;
            if(T[i] == 1 && (V[i] & 1) == 1 && C[i][1] == 0) return false;
            if(T[i] == 1 && (V[i] & 1) == 0 && C[i][1] != 0) return false;
        }
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                if(t[i][j]) ans[i][j] |= (1ull << k);
            }
        }
        for(int i = 1; i <= n; i++) {
            U[i] >>= 1;
            V[i] >>= 1;
        }
    }
    return true;
}
int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", &S[i]);
    for(int i = 1; i <= n; i++) scanf("%d", &T[i]);
    for(int i = 1; i <= n; i++) scanf("%llu", &U[i]);
    for(int i = 1; i <= n; i++) scanf("%llu", &V[i]);
    if(!solve()) printf("-1\n");
    else {
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                printf("%llu%c", ans[i][j], j == n ? '\n' : ' ');
            }
        }
    }
    return 0;
}