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;

const int N = 101;
const long long DMAX = 1e11;

long long g[N][N], mul[N];
int p[N];

int main() {
  mul[0] = 1;
  for (int i = 1; i < 10; ++i) {
    mul[i] = mul[i-1] + 1;
    bool ok = 0;
    while (!ok) {
      ok = 1;
      set<int> st;
      for (int j = 0; j <= i && ok; ++j) {
        if (st.count(mul[j])) {
          ok = 0;
          break;
        }
        st.insert(mul[j]);
        for (int k = j+1; k <= i; ++k) {
          int now = mul[k] + mul[j];
          if (st.count(now)) {
            ok = 0;
            break;
          }
          st.insert(now);
        }
      }
      if (ok) break;
      ++mul[i];
    }
  }
  int N, n = 2;
  scanf("%d", &N);
  g[0][1] = g[1][0] = 1;
  long long last = 1;
  while (n < N) {
    ++n;
    long long cur = last;
    bool ok = 0;
    while (!ok) {
      ok = 1;
      for (int i = 0; i < n-1; ++i)
        g[i][n-1] = g[n-1][i] = mul[i] * cur;
      last = 0;
      for (int i = 0; i < n; ++i)
        p[i] = i;
      set<long long> st;
      do {
        if (p[0] > p[n-1]) continue;
        long long sum = 0;
        for (int i = 0; i + 1 < n; ++i)
          sum += g[p[i]][p[i+1]];
        if (st.count(sum)) {
          ok = 0;
          break;
        }
        st.insert(sum);
        last = max(last, sum);
        ++cur;
      } while (next_permutation(p, p+n));
    }
  }
  assert(last <= DMAX);
  cerr << last << endl;
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < N; ++j) {
      printf("%lld%c", g[i][j], j == N-1 ? '\n' : ' ');
    }
  }
  return 0;
}