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: Wrong Answer)

//#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <queue>

using namespace std;

typedef long long ll;
typedef vector<int> vint;
typedef vector<vector<int> > vvint;
typedef vector<long long> vll, vLL;
typedef vector<vector<long long> > vvll, vvLL;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<n;++i)
#define mod (ll)(1e9+7)
#define FIX(a) ((a)%mod+mod)%mod
#define ALL(obj) (obj).begin(), (obj).end()
#define rALL(obj) (obj).rbegin(), (obj).rend()
#define INF 1000000000 //1e9
#define LLINF 2000000000000000000LL //2e18
#define fi first
#define se second
#define pb push_back
int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    int n2 = 1;
    while(n2<n){
        n2*=2;
    }
    vvint ed(n2+1,vint(n2+1,0));
    int st=1;
    int lv=1;
    while(st*2<=n2){
        if(st==1){
            int tm = 1;
            while(tm*2<n2){
                int now=0;
                ed[now][(now+tm)%n2]=1;
                ed[(now+tm)%n2][now]=lv;
                now=(now+tm)%n2;
                while(now!=0){
                    ed[now][(now+tm)%n2]=1;
                    ed[(now+tm)%n2][now]=lv;
                    now=(now+tm)%n2;
                }
                tm+=2;
            }
        }else{
            for(int i=0;i<st;i++){
                int now=i;
                ed[now][(now+st)%n2]=lv;
                ed[(now+st)%n2][now]=lv;
                now=(now+st)%n2;
                while(now!=i){
                    ed[now][(now+st)%n2]=lv;
                    ed[(now+st)%n2][now]=lv;
                    now=(now+st)%n2;
                }
            }
        }
        st*=2;
        lv++;
    }
    for(int i=0;i<n-1;i++){
        for(int j=i+1;j<n-1;j++){
            cout << ed[i][j] << " ";
        }
        cout << ed[i][n-1] << endl;
    }
    return 0;
}