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, s, i, j, k, t, a[100][100] = {0};
  cin >> n;
  t = (n * n) / 2;
  t++;
  k = 1;
  if (n % 2 != 0) {
    for (i = 0, j = n / 2; i < n; i++) {
      a[i][j] = k;
      k = k + 2;
      t--;
    }
    for (i = n / 2, j = 0; j < n; j++) {
      if (a[i][j] == 0) {
        a[i][j] = k;
        k = k + 2;
        t--;
      }
    }
    i = 0;
    while (t > 0) {
      for (i = 0; i < n / 2 && t > 0; i++) {
        for (j = i; j < n / 2 && t > 0; j++) {
          a[i][j] = k;
          k = k + 2;
          a[i][n - j - 1] = k;
          k = k + 2;
          a[n - i - 1][j] = k;
          k = k + 2;
          a[n - i - 1][n - j - 1] = k;
          k = k + 2;
          t -= 4;
        }
      }
    }
    k = 2;
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) {
        if (a[i][j] == 0) {
          a[i][j] = k;
          k += 2;
        }
      }
    }
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) cout << a[i][j] << " ";
      cout << endl;
    }
  }
  return 0;
}