Problem: Diverta City is a new city consisting of N towns numbered 1, 2, ..., N.

The mayor Ringo is planning to connect every pair of two different towns with a bidirectional road. The length of each road is undecided.

A Hamiltonian path is a path that starts at one of the towns and visits each of the other towns exactly once. The reversal of a Hamiltonian path is considered the same as the original Hamiltonian path.

There are N! / 2 Hamiltonian paths. Ringo wants all these paths to have distinct total lengths (the sum of the lengths of the roads on a path), to make the city diverse.

Find one such set of the lengths of the roads, under the following conditions:

* The length of each road must be a positive integer.
* The maximum total length of a Hamiltonian path must be at most 10^{11}.

Constraints

* N is a integer between 2 and 10 (inclusive).

Input

Input is given from Standard Input in the following format:


N


Output

Print a set of the lengths of the roads that meets the objective, in the following format:


w_{1, 1} \ w_{1, 2} \ w_{1, 3} \ ... \ w_{1, N}
w_{2, 1} \ w_{2, 2} \ w_{2, 3} \ ... \ w_{2, N}
:  :     :
w_{N, 1} \ w_{N, 2} \ w_{N, 3} \ ... \ w_{N, N}


where w_{i, j} is the length of the road connecting Town i and Town j, which must satisfy the following conditions:

* w_{i, i} = 0
* w_{i, j} = w_{j, i} \ (i \neq j)
* 1 \leq w_{i, j} \leq 10^{11} \ (i \neq j)



If there are multiple sets of lengths of the roads that meet the objective, any of them will be accepted.

Examples

Input

3


Output

0 6 15
6 0 21
15 21 0


Input

4


Output

0 111 157 193
111 0 224 239
157 224 0 258
193 239 258 0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

using ll=long long;
ll w[12][12];

void solve() {
    int n; cin >> n;
    vector<int> a = {1, 2, 4, 7, 12, 20, 29, 38, 52, 73};
    memset(w, 0, sizeof w);
    w[1][0] = w[0][1] = 1;
    ll M = 1;
    auto upd = [&](int x){
        vector<int> v(x);
        iota(v.begin(), v.end(), 0);
        ll mx = 0;
        do {
            ll sum = 0;
            for (int i = 1; i < x; i++) {
                sum += w[v[i-1]][v[i]];
            }
            mx = max(mx, sum);
        } while (next_permutation(v.begin(), v.end()));
        M = mx;
    };
    for (int i = 2; i < n; i++) {
        for (int j = 0; j < i; j++) {
            w[i][j] = w[j][i] = (M+1) * a[j];
        }
        upd(i+1);
        //cout << M << ' ';
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cout << w[i][j] << ' ';
        }
        cout << "\n";
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}