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"
#define MAXN 100009
#define INF 1000000007
#define mp(x,y) make_pair(x,y)
#define all(v) v.begin(),v.end()
#define pb(x) push_back(x)
#define wr cout<<"----------------"<<endl;
#define ppb() pop_back()
#define tr(ii,c) for(__typeof((c).begin()) ii=(c).begin();ii!=(c).end();ii++)
#define ff first
#define ss second
#define my_little_dodge 46
#define debug(x)  cerr<< #x <<" = "<< x<<endl;
using namespace std;
 
typedef long long ll;
typedef pair<int,int> PII;
template<class T>bool umin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
template<class T>bool umax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
int main(){
    int n;
    cin>>n;
    for(int i=1;i<n;i++){
    	for(int j=1;j+i<=n;j++){
  			if(n&1 and n>=5 and i==1 and j==n-1)
  				printf("3 ");
  			else	
    			printf("%d ",((j-1)&1)+1);
    	}
    	puts("");	
    }
	return 0;
}