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() {
  long long n, i, j, k = 0, l = 0, arr[51][51];
  vector<long long> v, p;
  cin >> n;
  for (i = 2; i < n * n; i = i + 2) {
    p.push_back(i);
  }
  for (i = 1; i <= n * n; i = i + 2) {
    v.push_back(i);
  }
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= n; j++) {
      arr[i][j] = -1;
    }
  }
  for (i = 1; i <= n / 2 + 1; i++) {
    for (j = 0; j < i; j++) {
      arr[i][n / 2 + 1 - j] = v[k];
      k++;
    }
    for (j = 1; j < i; j++) {
      arr[i][n / 2 + 1 + j] = v[k];
      k++;
    }
  }
  for (i = n; i > n / 2 + 1; i--) {
    for (j = 0; j < n - i + 1; j++) {
      arr[i][n / 2 + 1 - j] = v[k];
      k++;
    }
    for (j = 1; j < n - i + 1; j++) {
      arr[i][n / 2 + 1 + j] = v[k];
      k++;
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (arr[i][j] == -1) {
        arr[i][j] = p[l];
        l++;
      }
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      cout << arr[i][j] << " ";
    }
    cout << "\n";
  }
}