Educational Codeforces Round 16 C. Magic Odd Square

Find an n × n matrix with different numbers from 1 to n2,
so the sum in each row, column and both main diagonals are odd.

Input

The only line contains odd integer n (1 ≤ n ≤ 49).

Output

Print n lines with n integers.
All the integers should be different and from 1 to n2.
The sum in each row, column and both main diagonals should be odd.

Examples

input

1

output

1

input

3

output

2 1 4
3 5 7
6 9 8

给你一个奇数，让你构造一个n*n的矩阵，矩阵的每一行，每一列还有对角线上数字的和都是奇数。

因为和是奇数，所以每一行每一列还有对角线上的数字的奇数肯定是奇数个。我们用1表示奇数，2表示偶数画出两个矩阵来看一下

n=3时

2 1 2

1 1 12 1 2

n=5时

2 2 1 2 2

2 1 1 1 2

1 1 1 1 12 1 1 1 2

2 2 1 2 2 

可以发现到矩阵中心曼哈顿距离小于等于n/2的都是奇数，其他的都是偶数。（其实我是看别人的）

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

int main(void)
{
int n,i,j,x;
while(scanf("%d",&n)==1)
{
int odd = 1,even = 2;
int m = n/2 + 1;
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
if(abs(i - m) + abs(j - m) <= n/2)
{
x = odd;
odd+=2;
}
else
{
x = even;
even+=2;
}
printf("%d ",x);
}
printf("\n");
}
}
}