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;
bool a[55][55];
int odd = 1, even = 2, n, m, mid;
int nextOdd() {
  odd += 2;
  return odd - 2;
}
int nextEven() {
  even += 2;
  return even - 2;
}
int main() {
  cin >> n;
  mid = n / 2 + 1;
  for (int i = 1; i <= n; ++i) a[mid][i] = a[i][mid] = 1;
  for (int i = mid - 1; i >= 1; --i) {
    int cj1 = mid, ci = i, cj2 = mid;
    while (ci < mid) {
      a[ci][cj1] = a[ci][cj2] = 1;
      ++ci, --cj1, ++cj2;
    }
  }
  for (int i = mid + 1; i <= n; ++i) {
    int cj1 = mid, cj2 = mid, ci = i;
    while (ci > mid) {
      a[ci][cj1] = a[ci][cj2] = 1;
      --ci, --cj1, ++cj2;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (!a[i][j])
        cout << nextEven() << ' ';
      else
        cout << nextOdd() << ' ';
    }
    cout << '\n';
  }
  return 0;
}