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, a[100][100];
vector<int> odd, even;
bool Odd(int i, int j) {
  if (i * 2 == n + 1) return 1;
  if (i > n / 2) i = n - i + 1;
  int x = (i - 1) * 2 + 1;
  int l = 1, r = n;
  int now = n - x;
  while (now) {
    if (l == j || r == j) return 0;
    now -= 2;
    l++;
    r--;
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(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++) {
    for (int j = 1; j <= n; j++) {
      if (Odd(i, j))
        a[i][j] = odd.back(), odd.pop_back();
      else
        a[i][j] = even.back(), even.pop_back();
      cout << a[i][j] << " ";
    }
    cout << endl;
  }
  return 0;
}