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 i64 = long long;

int main() {
    int n;
    std::cin >> n;

    std::vector<i64> a { 1 };
    std::set<i64> s { 1 };
    for (int i = 1; i <= n; i++) {
        i64 x = a.back();
        bool ok = false;
        while (!ok) {
            x++;
            while (s.count(x)) x++;
            ok = true;
            for (int j = 0; j < i; j++) {
                if (s.count(a[j] + x)) {
                    ok = false;
                    break;
                }
            }
        }
        s.insert(x);
        for (int j = 0; j < i; j++) {
            s.insert(a[j] + x);
        }
        a.push_back(x);
    }

    // for (int i = 0; i < a.size(); i++) std::cout << a[i] << std::endl;
    std::vector<std::vector<i64>> ret(n, std::vector<i64>(n));

    i64 m = 0;
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < i; j++) {
            ret[i][j] = ret[j][i] = (m + 1) * a[j];
        }

        std::vector<int> p;
        for (int j = 0; j <= i; j++) p.push_back(j);
        do {
            i64 k = 0;
            for (int j = 0; j < i; j++) k += ret[p[j]][p[j + 1]];
            m = std::max(m, k);
        } while (std::next_permutation(p.begin(), p.end()));
    }

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) std::cout << ret[i][j] << " ";
        std::cout << std::endl;
    }

    return 0;
}