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.

Example

Input

3


Output

1 2
1
import java.util.*;
import java.lang.*;

class Mod{
    static final long MOD = 1000_000_007;
    static final long[] fact = makeFactorialArray(1000000);
    static final long[] factInv = makeInversedArray(fact);

    public static long add(long a, long b){
        long tmp = (a+b)%MOD;
        return tmp>=0 ? tmp : tmp+MOD;
    }
    public static long adds(long... ar){
        long ans=0;
        for(long x:ar) ans = Mod.add(ans, x);
        return ans;
    }
    public static long sub(long a, long b){
        return add(a,-b);
    }
    public static long mult(long a, long b){
        long tmp = (a*b)%MOD;
        return tmp>=0 ? tmp : tmp+MOD;
    }
    public static long mults(long... ar){
        long ans=1;
        for(long x:ar) ans = Mod.mult(ans, x);
        return ans;
    }
    public static long power(long a, long x){ //calculate a^x
        if(x<0)return 0;
        if(x==0)return 1;
        if(x%2==0) {
            long half = power(a,x/2);
            return mult(half,half);
        }
        return (a*power(a,x-1)) % MOD;
    }
    public static long inverse(long a){
        return power(a,MOD-2);
    }
    public static long div(long a, long b){
        return mult(a, inverse(b));
    }
    public static long[] makeFactorialArray(int size){
        long[] array = new long[size];
        array[0]=1;
        for(int i=1;i<size;i++){
            array[i]=mult(array[i-1],i);
        }
        return array;
    }
    public static long[] makeInversedArray(long[] original){
        long[] array = new long[original.length];
        for(int i=0;i<original.length;i++){
            array[i] = inverse(original[i]);
        }
        return array;
    }
    public static long combination(int n, int r){
        if(n<0 || r<0 || n<r) return 0;
        return mult(fact[n], mult(factInv[r],factInv[n-r]));
    }
    public static long gcd(long a, long b){
        if(a<b) return gcd(b,a);
        if(b==0) return a;
        if(a%b==0) return b;
        return gcd(b,a%b);
    }
    public static long lcm(long a, long b){
        return (a/gcd(a,b))*b;
    }

}

public class Main {
    static long firstDifBit(int x, int y){
        int z = x^y;
        for(int i=0; i<30; i++) if((z & (1<<i))!=0) return i;
        return -1;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        for(int i=0; i<N; i++){
            for(int j=i+1; j<N; j++){
                System.out.print(1+firstDifBit(i,j));
                if(j==N-1) System.out.println();
                else System.out.print(" ");
            }
        }
    }
}
