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)

// Test D 不确定正确性

#include <bits/stdc++.h>

using namespace std;

int n;
int c[501][501];

int main() {
	int n, cl = 0; cin >> n;
	if(n == 2) {
		puts("1");
		return 0;
	}
	for(int i = 1; i <= n; i += 2) {
		++cl;
		for(int j = i + 2; j <= n; j++)
			c[i][j] = c[i + 1][j] = cl;
	}
	for(int i = 2; i <= n; i += 2)
		c[i - 1][i] = c[i - 1][i + 1] % cl + 1;
	for(int i = 1; i < n; i++) {
		for(int j = i + 1; j <= n; j++)
			printf("%d ", c[i][j]);
		puts("");
	} return 0;
}