Averages in Place

Slavko is bored, so he fills an $N \times N$ table with positive integers. He is happy when all three conditions hold.

• The average of the numbers in each row is an integer, and that integer is in the same row.
• The average of the numbers in each column is an integer, and that integer is in the same column.
• All $N^2$ numbers in the table are different.

Find a table that makes Slavko happy.

The first line contains the integer $N$ ($1 \le N \le 100$).

Many tables satisfy the conditions, so this problem fixes one answer. Print exactly the table defined below. Every number in it is greater than $0$ and smaller than $10^9$.

• If $N = 2$, print $-1$ on the first line. No table satisfies the conditions for $N = 2$.
• If $N$ is odd, print $1$ through $N^2$ in row order, so the value in row $i$, column $j$ is $N(i-1)+j$.
• If $N$ is even and $N \ge 4$, define $s_1, s_2, \ldots, s_N$ by $s_k = k$ for every $k$ with $1 \le k \le N-1$, and $s_N = 3N/2$. The value in row $i$, column $j$ is $2N(s_i - 1) + s_j$.

Print the table on $N$ lines, each line holding $N$ integers separated by one space.

In the odd table the average of row $i$ is $N(i-1) + (N+1)/2$, which sits in the middle cell of row $i$. The average of column $j$ is $N(N-1)/2 + j$, which sits in the middle cell of column $j$. For $N = 3$ the row averages are $2$, $5$, $8$ and the column averages are $4$, $5$, $6$.

For even $N \ge 4$ the average of $s_1, s_2, \ldots, s_N$ is $N/2 + 1$, and $s_{N/2+1} = N/2 + 1$. That is why the row average lands in its row and the column average lands in its column. Since $2N$ is larger than the gap between the largest and the smallest $s_k$, the $N^2$ values are all different.

For $N = 2$ the average of two numbers equals one of them only when the two numbers are equal, and the conditions forbid equal numbers, so no table exists.