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 v, x[50][50], n;
vector<int> vec[2];
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) ++v, vec[v & 1].push_back(v);
  for (int i = 1; i <= n; ++i) x[i][(n + 1) >> 1] = 1;
  for (int i = 1; i <= n; ++i) x[(n + 1) >> 1][i] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if (x[i][j]) x[i][j] = vec[1].back(), vec[1].pop_back();
  for (int i = 1; i <= n and vec[1].size(); ++i)
    for (int j = 1; j <= n and vec[1].size(); ++j)
      if (x[i][j] == 0) {
        x[i][j] = vec[1].back(), vec[1].pop_back();
        x[n - i + 1][j] = vec[1].back(), vec[1].pop_back();
        x[i][n - j + 1] = vec[1].back(), vec[1].pop_back();
        x[n - i + 1][n - j + 1] = vec[1].back(), vec[1].pop_back();
      }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if (x[i][j] == 0) x[i][j] = vec[0].back(), vec[0].pop_back();
  for (int i = 1; i <= n; ++i, cout << '\n')
    for (int j = 1; j <= n; ++j, cout << ' ') cout << x[i][j];
  return 0;
}