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:
// And in the end, the love you take is equal to the love you make.
//                      Paul McCartney (the Beatles). The End. Abbey Road.

#include <bits/stdc++.h>
using namespace std;

using LL = long long;

namespace _buff {

const size_t BUFF = 1 << 19;
char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
char getc() {
  if (ib == ie) {
    ib = ibuf;
    ie = ibuf + fread(ibuf, 1, BUFF, stdin);
  }
  return ib == ie ? -1 : *ib++;
}

}

LL read() {
  using namespace _buff;
  LL ret = 0;
  bool pos = true;
  char c = getc();
  for (; (c < '0' || c > '9') && c != '-'; c = getc()) {
    assert(~c);
  }
  if (c == '-') {
    pos = false;
    c = getc();
  }
  for (; c >= '0' && c <= '9'; c = getc()) {
    ret = (ret << 3) + (ret << 1) + (c ^ 48);
  }
  return pos ? ret : -ret;
}

int main() {
  const vector<int> arr {1, 2, 4, 7, 12, 20, 29, 38, 52, 73};
  int n = read();
  vector<vector<LL> > w(n, vector<LL>(n));

  auto get_max = [&](int n) {
    vector<int> vc(n);
    iota(vc.begin(), vc.end(), 0);
    LL ret = 0;
    do {
      LL cur = 0;
      for (unsigned i = 1; i < vc.size(); ++i) {
        cur += w[vc[i - 1]][vc[i]];
      }
      ret = max(ret, cur);
    } while (next_permutation(vc.begin(), vc.end()));
    return ret;
  };

  LL m = 0;
  for (int i = 1; i < n; ++i) {
    for (int j = 0; j < i; ++j) {
      w[i][j] = w[j][i] = (m + 1) * arr[j];
    }
    m = get_max(i + 1);
  }

  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      cout << w[i][j] << "\n "[j < n - 1];
    }
  }
  return 0;
}