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() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  cin >> n;
  vector<int> odds;
  vector<int> evens;
  for (int i = 1; i <= n * n; ++i) {
    if (i % 2 == 0)
      evens.push_back(i);
    else
      odds.push_back(i);
  }
  vector<vector<int>> field(n);
  for (int i = 0; i < n; ++i) {
    field[i].resize(n, 0);
  }
  for (int i = 0; i <= n / 2; ++i) {
    int count = i * 2 + 1;
    while (count > 0) {
      field[i][n / 2 - count / 2] = 1;
      field[i][n / 2 + count / 2] = 1;
      count -= 2;
    }
  }
  int cnt = 1;
  for (int i = n / 2 + 1; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      field[i][j] = field[n / 2 - cnt][j];
    }
    ++cnt;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (field[i][j] == 0) {
        field[i][j] = evens.back();
        evens.pop_back();
      } else {
        field[i][j] = odds.back();
        odds.pop_back();
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      cout << field[i][j] << " ";
    }
    cout << endl;
  }
  return 0;
}