You are given a prime number $n$, and an array of $n$ integers $b_1,b_2,\ldots, b_n$, where $0 \leq b_i < n$ for each $1 \le i \leq n$.

You have to find a matrix $a$ of size $n \times n$ such that all of the following requirements hold:

$0 \le a_{i,j} < n$ for all $1 \le i, j \le n$.

$a_{r_1, c_1} + a_{r_2, c_2} \not\equiv a_{r_1, c_2} + a_{r_2, c_1} \pmod n$ for all positive integers $r_1$, $r_2$, $c_1$, and $c_2$ such that $1 \le r_1 < r_2 \le n$ and $1 \le c_1 < c_2 \le n$.

$a_{i,i} = b_i$ for all $1 \le i \leq n$.

Here $x \not \equiv y \pmod m$ denotes that $x$ and $y$ give different remainders when divided by $m$.

If there are multiple solutions, output any. It can be shown that such a matrix always exists under the given constraints.


-----Input-----

The first line contains a single positive integer $n$ ($2 \le n < 350$).

The second line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($0 \le b_i < n$) — the required values on the main diagonal of the matrix.

It is guaranteed that $n$ is prime.


-----Output-----

Print $n$ lines. On the $i$-th line, print $n$ integers $a_{i, 1}, a_{i, 2}, \ldots, a_{i, n}$, each separated with a space.

If there are multiple solutions, output any.


-----Examples-----

Input
2
0 0
Output
0 1 
0 0
Input
3
1 1 1
Output
1 2 2
1 1 0
1 0 1
Input
5
1 4 1 2 4
Output
1 0 1 3 4
1 4 3 1 0
2 4 1 0 2
1 2 2 2 2
2 2 0 1 4


-----Note-----

In the first example, the answer is valid because all entries are non-negative integers less than $n = 2$, and $a_{1,1}+a_{2,2} \not\equiv a_{1,2}+a_{2,1} \pmod 2$ (because $a_{1,1}+a_{2,2} = 0 + 0 \equiv 0 \pmod 2$ and $a_{1,2}+a_{2,1} = 1 + 0 \equiv 1 \pmod 2 $). Moreover, the values on the main diagonals are equal to $0,0$ as required.

In the second example, the answer is correct because all entries are non-negative integers less than $n = 3$, and the second condition is satisfied for all quadruplets $(r_1, r_2, c_1, c_2)$. For example:

When $r_1=1$, $r_2=2$, $c_1=1$ and $c_2=2$, $a_{1,1}+a_{2,2} \not\equiv a_{1,2}+a_{2,1} \pmod 3$ because $a_{1,1}+a_{2,2} = 1 + 1 \equiv 2 \pmod 3$ and $a_{1,2}+a_{2,1} = 2 + 1 \equiv 0 \pmod 3 $.

When $r_1=2$, $r_2=3$, $c_1=1$, and $c_2=3$, $a_{2,1}+a_{3,3} \not\equiv a_{2,3}+a_{3,1} \pmod 3$ because $a_{2,1}+a_{3,3} = 1 + 1 \equiv 2 \pmod 3$ and $a_{2,3}+a_{3,1} = 0 + 1 \equiv 1 \pmod 3 $.

Moreover, the values on the main diagonal are equal to $1,1,1$ as required.

Source: codeforces | Difficulty: HARD
Tags: Number theory

Constructive algorithms


Solutions:
n = int(input())
data = [[i * j % n for j in range(n)] for i in range(n)]
for (i, v) in enumerate(map(int, input().split())):
	delta = v - data[i][i]
	for j in range(n):
		data[i][j] += delta
		data[i][j] %= n
for a in data:
	print(' '.join((str(x) for x in a)))


import sys
input = sys.stdin.readline

def solve():
	n = int(input())
	b = list(map(int, input().split()))
	ans = [[0] * n for _ in range(n)]
	for i in range(n):
		ans[i][i] = b[i]
		j = i
		while j > 0:
			ans[i][j - 1] = (ans[i][j] - i) % n
			j -= 1
		j = i
		while j < n - 1:
			ans[i][j + 1] = (ans[i][j] + i) % n
			j += 1
	for row in ans:
		print(*row)
for _ in range(1):
	solve()


n = int(input())
a = list(map(int, input().split()))
for i in range(0, n):
	for j in range(0, n):
		print((((j - i) * i + a[i]) % n + n) % n, end=' ')
	print()


def main():
	m = int(input())
	l = list(map(int, input().split()))
	for k in range(m):
		for x in range(m):
			v = (((x - k) * k + l[k]) % m + m) % m
			print(v, end=' ')
		print()

def loop():
	t = 1
	for i in range(t):
		main()
loop()


n = int(input())
a = list(map(int, input().split()))
for i in range(0, n):
	ll = 0
	ll += 1
	for j in range(0, n):
		ok = 1
		ok -= 1
		ok += 1
		print((((j - i) * i + a[i]) % n + n) % n, end=' ')
		i = i + 1
		j = j + 1
		i -= 1
		j -= 1
	print()