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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<stdio.h>
#include<vector>
#include<queue>
#include<math.h>
#include<deque>
using namespace std;
#define int long long
#define rep(s,i,n) for(int i=s;i<n;i++)
#define c(n) cout<<n<<endl;
#define ic(n) int n;cin>>n;
#define sc(s) string s;cin>>s;
#define mod 1000000007
#define inf 1000000000000000007
#define f first
#define s second
#define mini(c,a,b) *min_element(c+a,c+b)
#define maxi(c,a,b) *max_element(c+a,c+b)
#define pi 3.141592653589793238462643383279
#define e_ 2.718281828459045235360287471352
#define P pair<int,int>
#define upp(a,n,x) upper_bound(a,a+n,x)-a;
#define low(a,n,x) lower_bound(a,a+n,x)-a;
#define UF UnionFind 
//printf("%.12Lf\n",);
int keta(int x) {
	rep(0, i, 30) {
		if (x < 10) {
			return i + 1;
		}
		x = x / 10;
	}
}
int gcd(int x, int y) {
	if (x == 0 || y == 0)return x + y;
	int aa = x, bb = y;
	rep(0, i, 1000) {
		aa = aa % bb;
		if (aa == 0) {
			return bb;
		}
		bb = bb % aa;
		if (bb == 0) {
			return aa;
		}
	}
}
int lcm(int x, int y) {
	int aa = x, bb = y;
	rep(0, i, 1000) {
		aa = aa % bb;
		if (aa == 0) {
			return x / bb * y;
		}
		bb = bb % aa;
		if (bb == 0) {
			return x / aa * y;
		}
	}
}
bool p(int x) {
	if (x == 1)return false;
	rep(2, i, sqrt(x) + 1) {
		if (x % i == 0 && x != i) {
			return false;
		}
	}
	return true;
}
int max(int a, int b) {
	if (a >= b)return a;
	else return b;
}
string maxst(string s, string t) {
	int n = s.size();
	int m = t.size();
	if (n > m)return s;
	else if (n < m)return t;
	else {
		rep(0, i, n) {
			if (s[i] > t[i])return s;
			if (s[i] < t[i])return t;
		}
		return s;
	}
}
int min(int a, int b) {
	if (a >= b)return b;
	else return a;
}
int n2[41];
int nis[41];
int nia[41];
int mody[41];
int nn;
int com(int n, int y) {
	int ni = 1;
	for (int i = 0;i < 41;i++) {
		n2[i] = ni;
		ni *= 2;
	}
	int bunsi = 1, bunbo = 1;
	rep(0, i, y)bunsi = (bunsi * (n - i)) % mod;
	rep(0, i, y)bunbo = (bunbo * (i + 1)) % mod;
	mody[0] = bunbo;
	rep(1, i, 41) {
		bunbo = (bunbo * bunbo) % mod;
		mody[i] = bunbo;
	}
	rep(0, i, 41)nis[i] = 0;
	nn = mod - 2;
	for (int i = 40;i >= 0;i -= 1) {
		if (nn > n2[i]) {
			nis[i]++;
			nn -= n2[i];
		}
	}
	nis[0]++;
	rep(0, i, 41) {
		if (nis[i] == 1) {
			bunsi = (bunsi * mody[i]) % mod;
		}
	}
	return bunsi;
}
int gyakugen(int n, int y) {
	int ni = 1;
	for (int i = 0;i < 41;i++) {
		n2[i] = ni;
		ni *= 2;
	}
	mody[0] = y;
	rep(1, i, 41) {
		y = (y * y) % mod;
		mody[i] = y;
	}
	rep(0, i, 41)nis[i] = 0;
	nn = mod - 2;
	for (int i = 40;i >= 0;i -= 1) {
		if (nn > n2[i]) {
			nis[i]++;
			nn -= n2[i];
		}
	}
	nis[0]++;
	rep(0, i, 41) {
		if (nis[i] == 1) {
			n = (n * mody[i]) % mod;
		}
	}
	return n;
}
int yakuwa(int n) {
	int sum = 0;
	rep(1, i, sqrt(n + 1)) {
		if (n % i == 0)sum += i + n / i;
		if (i * i == n)sum -= i;
	}
	return sum;
}
int poow(int y, int n) {
	if (n == 0)return 1;
	n -= 1;
	int ni = 1;
	for (int i = 0;i < 41;i++) {
		n2[i] = ni;
		ni *= 2;
	}
	int yy = y;
	mody[0] = yy;
	rep(1, i, 41) {
		yy = (yy * yy) % mod;
		mody[i] = yy;
	}
	rep(0, i, 41)nis[i] = 0;
	nn = n;
	for (int i = 40;i >= 0;i -= 1) {
		if (nn >= n2[i]) {
			nis[i]++;
			nn -= n2[i];
		}
	}
	rep(0, i, 41) {
		if (nis[i] == 1) {
			y = (y * mody[i]) % mod;
		}
	}
	return y;
}
int minpow(int x, int y) {
	int sum = 1;
	rep(0, i, y)sum *= x;
	return sum;
}
int ketawa(int x, int sinsuu) {
	int sum = 0;
	rep(0, i, 100)sum += (x % poow(sinsuu, i + 1)) / (poow(sinsuu, i));
	return sum;
}
int sankaku(int a) {
	return a * (a + 1) / 2;
}
int sames(int a[1111111], int n) {
	int ans = 0;
	rep(0, i, n) {
		if (a[i] == a[i + 1]) {
			int j = i;
			while (a[j + 1] == a[i] && j <= n - 2)j++;
			ans += sankaku(j - i);
			i = j;
		}
	}
	return ans;
}
using Graph = vector<vector<int>>;
int oya[114514];
int depth[114514];
void dfs(const Graph& G, int v, int p, int d) {
	depth[v] = d;
	oya[v] = p;
	for (auto nv : G[v]) {
		if (nv == p) continue; // nv が親 p だったらダメ
		dfs(G, nv, v, d + 1); // d を 1 増やして子ノードへ
	}
}
/*int H=10,W=10;
char field[10][10];
char memo[10][10];
void dfs(int h, int w) {
	memo[h][w] = 'x';

	// 八方向を探索
	for (int dh = -1; dh <= 1; ++dh) {
		for (int dw = -1; dw <= 1; ++dw) {
			if(abs(0-dh)+abs(0-dw)==2)continue;
			int nh = h + dh, nw = w + dw;

			// 場外アウトしたり、0 だったりはスルー
			if (nh < 0 || nh >= H || nw < 0 || nw >= W) continue;
			if (memo[nh][nw] == 'x') continue;

			// 再帰的に探索
			dfs(nh, nw);
		}
	}
}*/
int XOR(int a, int b) {
	if (a == 0 || b == 0) {
		return a + b;
	}
	int ni = 1;
	rep(0, i, 41) {
		n2[i] = ni;
		ni *= 2;
	}
	rep(0, i, 41)nis[i] = 0;
	for (int i = 40;i >= 0;i -= 1) {
		if (a >= n2[i]) {
			nis[i]++;
			a -= n2[i];
		}
		if (b >= n2[i]) {
			nis[i]++;
			b -= n2[i];
		}
	}
	int sum = 0;
	rep(0, i, 41)sum += (nis[i] % 2 * n2[i]);
	return sum;
}
//int ma[1024577][21];
//for(int bit=0;bit<(1<<n);bit++)rep(0,i,n)if(bit&(1<<i))ma[bit][i]=1;
struct UnionFind {
	vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
 
	UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
		for (int i = 0; i < N; i++) par[i] = i;
	}
 
	int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}
 
	void unite(int x, int y) { // xとyの木を併合
		int rx = root(x); //xの根をrx
		int ry = root(y); //yの根をry
		if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
		par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
	}
 
	bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}
};
int ans[514][514];
signed main(){
	ic(n) 
	rep(0,i,n)rep(0,j,n)ans[i][j]=0;
	rep(0,i,n/2){
		int l=i;
		int r=i+1;
		int b[514];
		b[0]=1;
		ans[l][r]=i+1;
		ans[r][l]=i+1;
		rep(1,j,(n-1)/2){
			ans[l][r]=i+1;
			ans[r][l]=i+1;
			if(j%2==0){
				l=r;
				r=(r+j+1)%n;
				b[j]=j+1;
			}
			else{
				l=r;
				r=(r-j-1+n)%n;
				b[j]=-(j+1);
			}	
		}	
		if(n%2==0){
			ans[r][l]=i+1;
			ans[l][r]=i+1;
			l=r;
			r=(r+n/2)%n;
		}
		rep(0,j,(n-1)/2){
			ans[r][l]=i+1;
			ans[l][r]=i+1;
			if(n%2==1){
				l=r;
				r=(r+b[(n-3)/2-j]+n)%n;
			}
			else{
				l=r;
				r=(r-b[(n-3)/2-j]+n)%n;
			}		
		}			
		ans[r][l]=i+1;
		ans[l][r]=i+1;
	}
	rep(0,i,n){
		if(n%2==1){
			ans[i][n-i-1]=(n+1)/2;
		}
	}
	rep(0,i,n){
		rep(i+1,j,n)cout<<ans[i][j]<<" ";
		cout<<endl;
	}	
	if(n>=7)c(n)
}