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:
//This Code was made by Chinese_zjc_.
#include <iostream>
#include <fstream>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <bitset>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <map>
#include <set>
#include <ctime>
// #include<windows.h>
#define int long long
#define double long double
using namespace std;
const double PI = acos(-1);
const double eps = 0.0000000001;
const int INF = 0x3fffffffffffffff;
int n, w[11][11] = {
           {},
           {0, 0, 1, 2, 7, 45, 512, 9925, 321381, 15870550, 1069877351},
           {0, 1, 0, 4, 14, 90, 1024, 19850, 642762, 31741100, 2139754702},
           {0, 2, 4, 0, 28, 180, 2048, 39700, 1285524, 63482200, 4279509404},
           {0, 7, 14, 28, 0, 315, 3584, 69475, 2249667, 111093850, 7489141457},
           {0, 45, 90, 180, 315, 0, 6144, 119100, 3856572, 190446600, 12838528212},
           {0, 512, 1024, 2048, 3584, 6144, 0, 198500, 6427620, 317411000, 21397547020},
           {0, 9925, 19850, 39700, 69475, 119100, 198500, 0, 9320049, 460245950, 31026443179},
           {0, 321381, 642762, 1285524, 2249667, 3856572, 6427620, 9320049, 0, 603080900, 40655339338},
           {0, 15870550, 31741100, 63482200, 111093850, 190446600, 317411000, 460245950, 603080900, 0, 56703499603},
           {0, 1069877351, 2139754702, 4279509404, 7489141457, 12838528212, 21397547020, 31026443179, 40655339338, 56703499603, 0},
};
signed main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    for (int i = 1; i <= n; ++i)
    {
        for (int j = 1; j <= n; ++j)
        {
            cout << w[i][j] << ' ';
        }
        cout << endl;
    }
    return 0;
}