Problem: Diverta City is a new city consisting of N towns numbered 1, 2, ..., N.

The mayor Ringo is planning to connect every pair of two different towns with a bidirectional road. The length of each road is undecided.

A Hamiltonian path is a path that starts at one of the towns and visits each of the other towns exactly once. The reversal of a Hamiltonian path is considered the same as the original Hamiltonian path.

There are N! / 2 Hamiltonian paths. Ringo wants all these paths to have distinct total lengths (the sum of the lengths of the roads on a path), to make the city diverse.

Find one such set of the lengths of the roads, under the following conditions:

* The length of each road must be a positive integer.
* The maximum total length of a Hamiltonian path must be at most 10^{11}.

Constraints

* N is a integer between 2 and 10 (inclusive).

Input

Input is given from Standard Input in the following format:


N


Output

Print a set of the lengths of the roads that meets the objective, in the following format:


w_{1, 1} \ w_{1, 2} \ w_{1, 3} \ ... \ w_{1, N}
w_{2, 1} \ w_{2, 2} \ w_{2, 3} \ ... \ w_{2, N}
:  :     :
w_{N, 1} \ w_{N, 2} \ w_{N, 3} \ ... \ w_{N, N}


where w_{i, j} is the length of the road connecting Town i and Town j, which must satisfy the following conditions:

* w_{i, i} = 0
* w_{i, j} = w_{j, i} \ (i \neq j)
* 1 \leq w_{i, j} \leq 10^{11} \ (i \neq j)



If there are multiple sets of lengths of the roads that meet the objective, any of them will be accepted.

Examples

Input

3


Output

0 6 15
6 0 21
15 21 0


Input

4


Output

0 111 157 193
111 0 224 239
157 224 0 258
193 239 258 0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

int n;
bool u[20];
map<long long, int> m;
vector<int> v;
long long cur = 1, x[20][20];

long long ma(int a, int b, long long c) {
	if (b == n)
		cur = max(cur, c);
	b++;
	for (int i = 0; i < n; i++)
		if (!u[i]) {
			u[i] = 1;
			ma(i, b, c + x[a][i]);
			u[i] = 0;
		}
	return 1;
}

bool check(int a, int b, long long c) {
	// printf("check(a=%d b=%d c=%lld)\n", a, b, c);
	if (b == n)
		return ++m[c] <= 2 || !printf("c=%lld\n", c);
	b++;
	for (int i = 0; i < n; i++)
		if (!u[i]) {
			u[i] = 1;
			bool ret = check(i, b, c + x[a][i]);
			u[i] = 0;
			if (!ret)
				return 0;
		}
	return 1;
}

int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
		for (int j = 1; ; j++) {
			bool ok = 1;
			set<int> s;
			v.push_back(j);
			for (int k : v)
				if (!s.insert(k).second)
					ok = 0;
			for (int k = 0; k < v.size(); k++)
				for (int l = k + 1; l < v.size(); l++)
					if (!s.insert(v[k] + v[l]).second)
						ok = 0;
			if (ok)
				break;
			else
				v.pop_back();
		}
	// printf("v="); for (int i : v) printf("%d ", i); puts("");
	x[0][1] = x[1][0] = 1;
	cur = 1;
	for (int i = 2; i < n; i++) {
		cur++;
		for (int j = 0; j < i; j++)
			x[i][j] = x[j][i] = cur * v[j];
		if (i == n - 1)
			break;
		for (int j = 0; j <= i; j++) {
			u[j] = 1;
			ma(j, 1, 0);
			u[j] = 0;
		}
	}
	/*
	printf("cur=%lld\n", cur);
	bool ok = 1;
	for (int i = 0; i < n && ok; i++) {
		u[i] = 1;
		ok &= check(i, 1, 0);
		u[i] = 0;
	}
	printf("ok=%d\n", ok);
	*/
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
			printf("%lld%c", x[i][j], " \n"[j == n - 1]);
}