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, ans[50][50];
vector<int> odd, even;
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = (1); i <= (n * n); i++) {
    if (i & 1)
      odd.push_back(i);
    else
      even.push_back(i);
  }
  for (int i = (1); i <= (n); i++) {
    ans[i][n / 2 + 1] = odd.back();
    odd.pop_back();
    if (i <= n / 2) {
      for (int ii = (1); ii <= (i - 1); ii++) {
        ans[i][n / 2 + 1 + ii] = odd.back();
        odd.pop_back();
        ans[i][n / 2 + 1 - ii] = odd.back();
        odd.pop_back();
      }
    } else {
      for (int ii = (1); ii <= (n - i); ii++) {
        ans[i][n / 2 + 1 + ii] = odd.back();
        odd.pop_back();
        ans[i][n / 2 + 1 - ii] = odd.back();
        odd.pop_back();
      }
    }
    for (int j = (1); j <= (n); j++)
      if (!ans[i][j]) {
        ans[i][j] = even.back();
        even.pop_back();
      }
    for (int j = (1); j <= (n); j++) cout << ans[i][j] << ' ';
    cout << endl;
  }
}