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;
long long a[55][55];
int main() {
  memset(a, 0, sizeof(a));
  int n;
  cin >> n;
  int cnt = n * n / 2 + 1;
  for (int i = 1; i <= n; i++) {
    if (i == n / 2 || i == n / 2 + 2)
      ;
    else {
      a[i][i] = 1;
      a[i][n - i + 1] = 1;
      if (i == n / 2 + 1)
        cnt -= 1;
      else
        cnt -= 2;
    }
  }
  for (int i = 1; i <= n; i++) {
    a[i][n / 2 + 1] = 1;
    a[n / 2 + 1][i] = 1;
  }
  cnt -= 2 * n - 2;
  while (cnt > 1) {
    cnt -= 4;
    bool flag = 0;
    for (int i = 1; i <= n; i++) {
      if (flag) break;
      for (int j = 1; j <= n; j++) {
        if ((i == j) || (i + j == n + 1) || (i == n / 2 + 1) ||
            (j == n / 2 + 1))
          continue;
        if (a[i][j] == 0) {
          a[i][j] = 1;
          a[i][n + 1 - j] = 1;
          a[n + 1 - i][j] = 1;
          a[n + 1 - i][n + 1 - j] = 1;
          flag = 1;
          break;
        }
      }
    }
  }
  int x = 2, y = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (a[i][j] == 0) {
        cout << x << ' ';
        x += 2;
      } else {
        cout << y << ' ';
        y += 2;
      }
    }
    cout << endl;
  }
  return 0;
}