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 <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
#include <fstream>
#define popcount __builtin_popcount
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;

int main()
{
    int n; cin>>n;

    ll a[12];
    a[0]=1, a[1]=2;
    set<ll> st;
    st.insert(1); st.insert(2); st.insert(3);
    for(int i=2; i<=n; i++){
        for(a[i]=a[i-1]+1; ; a[i]++){
            if(st.find(a[i])!=st.end()) continue;
            bool ok=1;
            for(int j=0; j<i; j++){
                if(st.find(a[j]+a[i])!=st.end()){
                    ok=0; break;
                }
            }
            if(ok){
                st.insert(a[i]);
                for(int j=0; j<i; j++) st.insert(a[j]+a[i]);
                break;
            }
        }
    }

    ll w[12][12]={};
    w[0][1]=w[1][0]=1, w[0][2]=w[2][0]=2, w[1][2]=w[2][1]=3;
    ll m=5;
    for(int i=3; i<n; i++){
        for(int j=0; j<i; j++) w[j][i]=w[i][j]=(m+1)*a[j];
        vector<int> v(i+1);
        for(int j=0; j<=i; j++) v[j]=j;
        while(1){
            ll s=0;
            for(int j=0; j<i; j++){
                s+=w[v[j]][v[j+1]];
            }
            m=max(m, s);
            if(!next_permutation(v.begin(), v.end())) break;
        }
    }
    
    //cerr<<m<<endl;
    for(int i=0; i<n; i++){
        for(int j=0; j<n; j++){
            cout<<w[i][j]<<" ";
        }
        cout<<endl;
    }
    return 0;
}