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 even[(n * n) / 2], odd[(n * n) / 2 + 1], j = 0, k = 0;
  for (int i = 1; i <= n * n; i += 2) {
    odd[j++] = i;
    even[k++] = i + 1;
  }
  odd[j] = n * n;
  int p = 0, q = 0, t = 0, i = 0;
  int a[n][n], no = 1, ne;
  for (int j = 0; j < n; j++) {
    if (j == 0)
      no = 1;
    else if (t == 0) {
      no += 2;
      if (no == n) t = 1;
    } else
      no -= 2;
    i = 0;
    ne = (n - no) / 2;
    for (; i < ne; i++) a[i][j] = even[p++];
    for (int k = 0; k < no; k++, i++) a[i][j] = odd[q++];
    for (int k = 0; k < ne; k++, i++) a[i][j] = even[p++];
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) cout << a[i][j] << " ";
    cout << "\n";
  }
}