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>
//ios::sync_with_stdio(false);
//cin.tie(0);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<pii,int> ppii;
typedef pair<int,pii> pipi;
typedef pair<ll,ll> pll;
typedef pair<pll,ll> ppll;
typedef pair<ll,pll> plpl;
typedef tuple<ll,ll,ll> tl;
ll mod=1000000007;
ll mod2=998244353;
ll mod3=1000003;
ll mod4=998244853;
ll inf=1000000000000000000;
double pi=2*acos(0);
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,n,m) for(ll i=n;i>=m;i--)
int dh[4]={1,-1,0,0};
int dw[4]={0,0,1,-1};
int ddh[8]={-1,-1,-1,0,0,1,1,1};
int ddw[8]={-1,0,1,-1,1,-1,0,1};
ll lmax(ll a,ll b){
    if(a<b)return b;
    else return a;
}
ll lmin(ll a,ll b){
    if(a<b)return a;
    else return b;
}
ll gcd(ll a,ll b){
    if(a<b)swap(a,b);
    if(a%b==0)return b;
    return gcd(b,a%b);
}
ll Pow(ll n,ll k){
    ll ret=1;
    ll now=n;
    while(k>0){
        if(k&1)ret*=now;
        now*=now;
        k/=2;
    }
    return ret;
}
ll gya[200010];
ll kai[200010];
ll beki(ll n,ll k){
    ll ret=1;
    ll now=n;
    while(k>0){
        if(k%2==1){
            ret*=now;
            ret%=mod;
        }
        now*=now;
        now%=mod;
        k/=2;
    }
    return ret;
}
ll gyaku(ll n){
    return beki(n,mod-2);
}
void nckinit(ll n){
    kai[0]=1;
    kai[1]=1;
    for(int i=2;i<=n;i++){
        kai[i]=kai[i-1]*i;
        kai[i]%=mod;
    }
    gya[n]=gyaku(kai[n]);
    for(int i=n-1;i>=1;i--){
        gya[i]=gya[i+1]*(i+1);
        gya[i]%=mod;
    }
    gya[0]=1;
}
ll nck(ll n,ll k){
    if(k<0)return 0;
    if(k==0||n==k)return 1;
    ll ret=kai[n];
    ret*=gya[n-k];
    ret%=mod;
    ret*=gya[k];
    ret%=mod;
    return ret;
}
ll npk(ll n,ll k){
    if(k<0)return 0;
    if(k==0)return 1;
    ll ret=kai[n];
    ret*=gya[n-k];
    ret%=mod;
    return ret;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;cin>>n;
    int a[n+1][n+1];
    rep(i,0,n+1)rep(j,0,n+1)a[i][j]=0;
    rep(i,1,n)a[i][i+1]=1;
    if(n%2==1)a[1][n]=2;
    else a[1][n]=1;
    int now=2;
    int cnt=3;
    rep(i,2,n){
        if(cnt==i){
            now++;
            cnt+=now;
        }
        rep(j,1,i+1){
            for(int k=j;k<=n;k+=i){
                if(k+i>n)break;
                a[k][k+i]=now;
            }
        }
    }
    rep(i,1,n){
        rep(j,i+1,n+1)cout<<a[i][j]<<" ";
        cout<<endl;
    }
}