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 m;
int f[55][55];
int n;
int main() {
  scanf("%d", &m);
  int mid = m / 2 + 1;
  for (int i = 1; i <= m; i++) f[i][mid] = f[mid][i] = 1;
  int now = m * 2 - 1;
  int need = m * m / 2 + 1 - now;
  need /= 4;
  now = need;
  need = now;
  for (int i = 1; i < mid; i++)
    for (int j = 1; j < mid; j++)
      if (need != 0) {
        f[i][j] = 1;
        need--;
      }
  need = now;
  for (int i = 1; i < mid; i++)
    for (int j = m; j > mid; j--)
      if (need != 0) {
        f[i][j] = 1;
        need--;
      }
  need = now;
  for (int i = m; i > mid; i--)
    for (int j = 1; j < mid; j++)
      if (need != 0) {
        f[i][j] = 1;
        need--;
      }
  need = now;
  for (int i = m; i > mid; i--)
    for (int j = m; j > mid; j--)
      if (need != 0) {
        f[i][j] = 1;
        need--;
      }
  int odd = 1, even = 2;
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= m; j++) {
      if (f[i][j] == 1) {
        f[i][j] = odd;
        odd += 2;
      } else {
        f[i][j] = even;
        even += 2;
      }
      cout << f[i][j] << " ";
    }
    cout << endl;
  }
  return 0;
}