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;
  vector<vector<int> > a(n, vector<int>(n));
  for (int i = 0; i < n; ++i) {
    a[n / 2][i] = a[i][n / 2] = 1;
  }
  int put = 2 * n - 1;
  int req = (n * n + 1) / 2;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (put < req && !a[i][j]) {
        a[i][j] = a[n - 1 - i][j] = a[i][n - 1 - j] = a[n - 1 - i][n - 1 - j] =
            1;
        put += 4;
      }
    }
  }
  vector<int> odd, even;
  for (int i = 1; i <= n * n; ++i) {
    if (i & 1) {
      odd.push_back(i);
    } else {
      even.push_back(i);
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (a[i][j]) {
        cout << odd.back();
        odd.pop_back();
      } else {
        cout << even.back();
        even.pop_back();
      }
      cout << " ";
    }
    cout << "\n";
  }
}