Description

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.

2
0 0

3
1 1 1

5
1 4 1 2 4

0 1 
0 0

1 2 2
1 1 0
1 0 1

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.