Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int arr[50][50];
int main() {
  int n;
  int k1 = 1;
  int k2 = 2;
  scanf("%d", &n);
  for (int i = 1; i <= (n + 1) / 2; ++i) {
    for (int j = 1; j <= (n + 1) / 2 - i; ++j) {
      arr[i][j] = k2;
      k2 += 2;
    }
    for (int j = (n + 1) / 2 - i + 1; j <= (n + 1) / 2 + i - 1; ++j) {
      arr[i][j] = k1;
      k1 += 2;
    }
    for (int j = (n + 1) / 2 + i; j <= n; ++j) {
      arr[i][j] = k2;
      k2 += 2;
    }
  }
  for (int i = (n + 1) / 2 + 1; i <= n; ++i) {
    int t = n - i + 1;
    for (int j = 1; j <= (n + 1) / 2 - t; ++j) {
      arr[i][j] = k2;
      k2 += 2;
    }
    for (int j = (n + 1) / 2 - t + 1; j <= (n + 1) / 2 + t - 1; ++j) {
      arr[i][j] = k1;
      k1 += 2;
    }
    for (int j = (n + 1) / 2 + t; j <= n; ++j) {
      arr[i][j] = k2;
      k2 += 2;
    }
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) {
      printf("%d", arr[i][j]);
      if (j != n)
        putchar(' ');
      else
        putchar(10);
    }
  return 0;
}