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)

//2019.8.24 by ljz
//email 573902690@qq.com
//if you find any bug in my code
//please tell me
#include<bits/stdc++.h>
using namespace std;
#define res register int
#define LL long long
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f
#define unl __int128
#define eps 5.6e-8
#define RG register
#define db double
#define pc(x) __builtin_popcount(x)
typedef pair<int,int> Pair;
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pb push_back
#define ull unsigned LL
#define gc getchar
//inline char gc() {
//    static char buf[100000],*p1,*p2;
//    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
//}
inline int read() {
    res s=0,ch=gc();
    while(ch<'0'||ch>'9')ch=gc();
    while(ch>='0'&&ch<='9')s=s*10+ch-'0',ch=gc();
    return s;
}
//inline int read() {
//    res s=0,ch=gc(),w=1;
//    while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=gc();}
//    while(ch>='0'&&ch<='9')s=s*10+ch-'0',ch=gc();
//    return s*w;
//}
//inline LL Read() {
//    RG LL s=0;
//    res ch=gc();
//    while(ch<'0'||ch>'9')ch=gc();
//    while(ch>='0'&&ch<='9')s=s*10+ch-'0',ch=gc();
//    return s;
//}
//inline LL Read() {
//    RG LL s=0;
//    res ch=gc(),w=1;
//    while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=gc();}
//    while(ch>='0'&&ch<='9')s=s*10+ch-'0',ch=gc();
//    return s*w;
//}
//inline void write(RG unl x){
//    if(x>10)write(x/10);
//    putchar(int(x%10)+'0');
//}
inline void swap(res &x,res &y) {
    x^=y^=x^=y;
}
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
//clock_t start=clock();
//inline void ck(){
//    if(1.0*(clock()-start)/CLOCKS_PER_SEC>0.1)exit(0);
//}
const int N=500+10;
const int kcz=1e9+7;
namespace MAIN{
    int n;
    inline void add(res &x,const res &y){
        x+=y,x>=kcz?x-=kcz:1;
    }
    inline int mul(const res &x,const res &y){
        return int(1LL*x*y%kcz);
    }
    inline int Add(const res &x,const res &y){
        return x+y>=kcz?x+y-kcz:x+y;
    }
    int a[N][N];
    inline void solve(res x,res y,res tot){
        if(x==y)return;
        res p=(x+y+1)/2;
        for(res i=x;i<p;i++)
            for(res j=p;j<=y;j++)
                a[i][j]=tot;
        solve(x,p-1,tot+1),solve(p,y,tot+1);
    }
    inline void MAIN(){
        n=read(),solve(1,n,1);
        for(res i=1;i<=n;i++){
            for(res j=i+1;j<=n;j++)
                printf("%d ",a[i][j]);
            puts("");
        }
    }
}
int main(){
//    srand(19260817);
//    freopen("signin.in","r",stdin);
//    freopen("signin.out","w",stdout);
    MAIN::MAIN();
    return 0;
}