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 m[64][64];
void solve() {
  cin >> n;
  int p = 1, q = 2;
  memset(m, 0, sizeof(m));
  for (int i = 0; i <= n / 2; i++)
    for (int j = n / 2 - i; j < n - (n / 2 - i); j++) m[i][j] = p, p += 2;
  for (int i = n / 2 + 1; i < n; i++)
    for (int j = i - n / 2; j < n - (i - n / 2); j++) m[i][j] = p, p += 2;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (m[i][j] == 0) m[i][j] = q, q += 2;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) cout << m[i][j] << " ";
    cout << endl;
  }
}
bool check() {
  for (int i = 0; i < n; i++) {
    int sum = 0;
    for (int j = 0; j < n; j++) sum += m[i][j];
    if (sum & 1 != 1) return false;
    sum = 0;
    for (int j = 0; j < n; j++) sum += m[j][i];
    if (sum & 1 != 1) return false;
  }
  int sum = 0;
  for (int i = 0; i < n; i++) sum += m[i][i];
  if (sum & 1 != 1) return false;
  sum = 0;
  for (int i = 0; i < n; i++) sum += m[i][n - 1 - i];
  if (sum & 1 != 1) return false;
  return true;
}
int main() {
  solve();
  return 0;
}