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 <algorithm>
#include <bitset>
#include <cassert>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef pair<pii, int> ppiii;

int scale[10]{1,2,4,7,12,20,29,38,52,73};

ll dp[10][10];

void dfs(int curr, int mask, int v, ll currW, ll& ret) {
  if(__builtin_popcount(mask) == v) {
    ret = max(ret, currW);
    return;
  }
  for(int i = 0; i < v; i++) {
    if((mask&(1<<i)) == 0) {
      dfs(i, mask | (1<<i), v, currW + dp[curr][i], ret);
    }
  }
}

void solve() {
  int n;
  cin >> n;
  dp[0][1] = 1;
  dp[1][0] = 1;
  for(int v = 2; v < n; v++) {
    ll ret = 0;
    for(int i = 0; i < v; i++) {
      dfs(i, 1<<i, v, 0, ret);
    }
    for(int k = 0; k < v; k++) {
      dp[v][k] = (ret+1) * scale[k];
      dp[k][v] = (ret+1) * scale[k];
    }
  }
  for(int i = 0; i < n; i++) {
    for(int j = 0; j < n; j++) {
      cout << dp[i][j];
      if(j == n-1) cout << "\n";
      else cout << " ";
    }
  }
}

void casesolve() {
  int t;
  cin >> t;
  for(int i = 1; i <= t; i++) {
    cout << "Case #" << i << ": ";
    solve();
  }
}

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