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 n;
int matrix[55][55];
int main() {
  while (scanf("%d", &n) != EOF) {
    if (n == 1) {
      cout << 1 << endl;
      continue;
    }
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) matrix[i][j] = 1;
    int cnt = n * n / 2;
    for (int i = 1; i <= n / 2; i++) {
      for (int j = 1; j <= n / 2; j++) {
        matrix[i][j] = matrix[n + 1 - i][j] = matrix[i][n + 1 - j] =
            matrix[n + 1 - i][n + 1 - j] = 0;
        cnt -= 4;
        if (cnt == 0) break;
      }
      if (cnt == 0) break;
    }
    int odd = 1, even = 2;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        if (matrix[i][j]) {
          matrix[i][j] = odd;
          odd += 2;
        } else {
          matrix[i][j] = even;
          even += 2;
        }
        printf("%d", matrix[i][j]);
        if (j != n) printf(" ");
      }
      printf("\n");
    }
  }
  return 0;
}