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;
vector<int> odd;
vector<int> even;
int arr[100][100];
int main() {
  int n;
  cin >> n;
  if (n == 1) {
    cout << 1 << endl;
    return 0;
  }
  int cur = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      arr[i][j] = 0;
      if (cur % 2)
        odd.push_back(cur);
      else
        even.push_back(cur);
      cur++;
    }
  }
  int mid = n / 2 + 1;
  for (int j = 1; j <= n; j++) {
    arr[mid][j] = odd[0];
    odd.erase(odd.begin());
  }
  int x = mid - 1;
  int y = mid;
  int t = mid - 1;
  int cnt = 1;
  while (t--) {
    for (int i = y; i < y + cnt; i++) {
      arr[x][i] = odd[0];
      odd.erase(odd.begin());
    }
    y--;
    x--;
    cnt = cnt + 2;
  }
  x = mid + 1;
  y = mid;
  t = mid - 1;
  cnt = 1;
  while (t--) {
    for (int i = y; i < y + cnt; i++) {
      arr[x][i] = odd[0];
      odd.erase(odd.begin());
    }
    y--;
    x++;
    cnt = cnt + 2;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (arr[i][j] == 0) {
        arr[i][j] = even[0];
        even.erase(even.begin());
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cout << arr[i][j] << " ";
    }
    cout << endl;
  }
}