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 <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int N = 200000 + 5, mod = 1000000007;
inline int add (int u, int v) { return u + v >= mod ? u + v - mod : u + v; }
inline int mns (int u, int v) { return u - v < 0 ? u - v + mod : u - v; }
inline int mul (int u, int v) { return 1LL * u * v % mod; }
inline int qpow (int u, int v) {
	int tot = 1, base = u % mod;
	while (v) {
		if (v & 1) tot = mul (tot, base);
		base = mul (base, base);
		v >>= 1;
	}
	return tot;
}

int mp[505][505], n;

int main () {
	scanf ("%d", &n);
	int cnt = 0;
	for (int i = n; i >= 1; --i) {
		++cnt;
		for (int j = 1; j < i; ++j) {
			mp[i][j] = mp[j][i] = cnt;
		}
	}
	for (int i = 1; i <= n - 1; ++i) {
		for (int j = i + 1; j <= n; ++j) {
			printf ("%d ", mp[i][j]);
		}
		puts ("");
	}
	return 0;
}