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;
void solve() {
  int n;
  cin >> n;
  vector<vector<int>> a(n, vector<int>(n, 0));
  for (int i = 0; i < n / 2; i++) {
    for (int j = n / 2; j < n / 2 + i + 1; j++) {
      a[i][j] = 1;
    }
    for (int j = n / 2; j > n / 2 - i - 1; j--) {
      a[i][j] = 1;
    }
  }
  reverse(a.begin(), a.end());
  for (int i = 0; i < n / 2; i++) {
    for (int j = n / 2; j < n / 2 + i + 1; j++) {
      a[i][j] = 1;
    }
    for (int j = n / 2; j > n / 2 - i - 1; j--) {
      a[i][j] = 1;
    }
  }
  for (int i = 0; i < n; i++) {
    a[n / 2][i] = 1;
  }
  set<int> q0, q1;
  for (int i = 0; i < n * n; i++) {
    if ((i + 1) % 2 == 0) {
      q0.insert(i + 1);
    } else {
      q1.insert(i + 1);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (a[i][j]) {
        cout << *q1.begin() << " ";
        q1.erase(q1.begin());
      } else {
        cout << *q0.begin() << " ";
        q0.erase(q0.begin());
      }
    }
    cout << "\n";
  }
}
int32_t main() {
  int t = 1;
  while (t--) {
    solve();
  }
}