Problem:
Score: 600 points

Problem Statement

AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.

For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:

For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.

Your task is to set levels to the passages so that the highest level of a passage is minimized.

Constraints

N is an integer between 2 and 500 (inclusive).

Input

Input is given from Standard Input in the following format:

N

Output

Print one way to set levels to the passages so that the objective is achieved, as follows:

a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}

Here a_{i,j} is the level of the passage connecting Room i and Room j.

If there are multiple solutions, any of them will be accepted.

Sample Input 1

3

Sample Output 1

1 2
1

The following image describes this output:

For example, if we leave Room 2, traverse the path 2 \to 3 \to 2 \to 3 \to 2 \to 1 \to 2 while only passing passages of level 1 and get back to Room 2, we pass through a passage six times.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define FOR(i,n,m) for(int i=(n);i<(m);i++)
#define REP(i,n) for(int i=0;i<(n);i++)
#define REPR(i,n) for(int i=(n);i>=0;i--)
#define all(vec) vec.begin(),vec.end()
using vi=vector<int>;
using vvi=vector<vi>;
using vl=vector<ll>;
using vvl=vector<vl>;
using P=pair<ll,ll>;
using PP=pair<ll,P>;
using vp=vector<P>;
using vpp=vector<PP>;
using vs=vector<string>;
#define fi first
#define se second
#define pb push_back
template<class T>bool chmax(T &a,const T &b){if(a<b){a=b;return true;}return false;}
template<class T>bool chmin(T &a,const T &b){if(a>b){a=b;return true;}return false;}
template<typename A,typename B>istream&operator>>(istream&is,pair<A,B> &p){is>>p.fi>>p.se;return is;}
template<typename A,typename B>ostream&operator<<(ostream&os,const pair<A,B> &p){os<<"("<<p.fi<<","<<p.se<<")";return os;}
template<typename T>istream&operator>>(istream&is,vector<T> &t){REP(i,t.size())is>>t[i];return is;}
template<typename T>ostream&operator<<(ostream&os,const vector<T>&t){os<<"{";REP(i,t.size()){if(i)os<<",";os<<t[i];}cout<<"}";return os;}
const ll MOD=1000000007LL;
const int INF=1<<30;
const ll LINF=1LL<<60;
int dfs(vector<vector<int>>&ans,int l,int r,int x=1){
    //[l,r)を塗る
    if(r-l==1){
        return 0;
    }
    int nx=0;
    chmax(nx,dfs(ans,l,(l+r+1)/2,x));
    chmax(nx,dfs(ans,(l+r+1)/2,r,x));
    x=nx+1;
    FOR(i,l,(l+r+1)/2){
        FOR(j,(l+r+1)/2,r){
            ans[i][j]=ans[j][i]=x;
        }
    }
    return x;
}
int main(){
    int n;
    cin>>n;
    vvi ans(n,vi(n));
    dfs(ans,0,n);
    REP(i,n){
        bool f=false;
        REP(j,n){
            if(i<j){
                if(f)cout<<" ";
                cout<<ans[i][j];
                f=true;
            }
        }cout<<endl;
    }
    return 0;
}