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 main() {
  int n;
  cin >> n;
  int arr[n][n];
  int even = 2;
  int odd = 1;
  int pos = 0;
  int row = 0;
  memset(arr, 0, sizeof(arr));
  for (int i = 0; i <= n / 2; i++) {
    pos = 0;
    for (int j = i; j < n / 2; j++) pos++;
    for (int k = 0; k < 2 * i + 1; k++) {
      arr[row][pos] = odd;
      odd = odd + 2;
      pos++;
    }
    row++;
  }
  for (int i = n / 2; i > 0; i--) {
    pos = 0;
    for (int j = i; j <= n / 2; j++) {
      pos++;
    }
    for (int k = 0; k < 2 * i - 1; k++) {
      arr[row][pos] = odd;
      odd = odd + 2;
      pos++;
    }
    row++;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (!arr[i][j]) {
        arr[i][j] = even;
        even = even + 2;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      printf("%d ", arr[i][j]);
    }
    printf("\n");
  }
  return 0;
}