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 <iostream>
#include <vector>
#include <cstring>
using namespace std;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

const long long w[10] = {1, 2, 4, 7, 12, 20, 29, 38, 52, 73};

long long dp[(1<<10)+1][11];
long long calc(vector<vector<long long> > &G, int N) {
    memset(dp, 0, sizeof(dp));
    for (int bit = 0; bit < (1<<N); ++bit) {
        for (int v = 0; v < N; ++v) {
            if (!(bit & (1<<v))) continue;
            for (int nv = 0; nv < N; ++nv) {
                if (bit & (1<<nv)) continue;
                int nbit = bit | (1<<nv);
                chmax(dp[nbit][nv], dp[bit][v] + G[v][nv]);
            }
        }
    }
    long long res = 0;
    for (int v = 0; v < N; ++v) chmax(res, dp[(1<<N)-1][v]);
    return res;
}

int main() {
    int N; cin >> N;
    vector<vector<long long> > G(N, vector<long long>(N, 0));

    G[0][1] = 1; G[1][0] = 1;
    for (int n = 2; n < N; ++n) {
        long long M = calc(G, n);
        for (int v = 0; v < n; ++v) G[v][n] = G[n][v] = (M+1) * w[v];

        //cout << n << ": " << M << endl;
    }

    //cout << calc(G, N) << endl;
    
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) cout << G[i][j] << " ";
        cout << endl;
    }
}