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 <algorithm>
#include <set>

using namespace std;

vector<int> gen(int N){
    vector<int> res;
    set<int> S;
    int last = 1;
    for(int i=0;i<N;i++){
        while(true){
            bool ok = !S.count(last);
            if(ok){
                for(auto& t : res){
                    if(S.count(t + last)){
                        ok = false;
                        break;
                    }
                }
            }
            if(ok){
                S.insert(last);
                for(auto& t : res) S.insert(t + last);
                res.push_back(last);
                ++last;
                break;
            }
            ++last;
        }
    }
    return res;
}

long long longest(const vector<vector<long long>>& cost, int N){
    vector<int> a(N);
    for(int i=0;i<N;i++) a[i] = i;
    long long res = 1;
    do {
        long long sum = 0;
        for(int i=0;i+1<N;i++) sum += cost[a[i]][a[i+1]];
        res = max(res, sum);
    } while(next_permutation(a.begin(), a.end()));
    return res;
}

vector<vector<long long>> solve(int N){
    vector<int> g = gen(N);
    vector<vector<long long>> res(N, vector<long long>(N, 0));
    res[0][1] = res[1][0] = 1;
    for(int i=2;i<N;i++){
        long long l = longest(res, i);
        for(int j=0;j<i;j++) res[i][j] = res[j][i] = g[j] * (l+1);
    }
    return res;
}

int main(){
    int N;
    while(cin >> N){
        auto l = solve(N);
        for(auto& v : l){
            cout << v[0];
            for(int i=1;i<v.size();i++) cout << " " << v[i];
            cout << endl;
        }
    }
}