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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll loops")
using namespace std;
const long long int MOD = 998244353;
const long long int INF = 1e18;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  vector<vector<long long int>> a(n, vector<long long int>(n, 1));
  long long int o = 1;
  long long int e = 2;
  long long int p = (n * n + 1) / 2;
  long long int q = n * n - p;
  for (long long int i = 0; i < n / 2; i++) {
    for (long long int j = 0; j < n / 2; j++) {
      if (q > 0) {
        q -= 4;
        a[i][j] = 0;
        a[n - 1 - i][j] = 0;
        a[n - 1 - i][n - 1 - j] = 0;
        a[i][n - 1 - j] = 0;
      }
    }
  }
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < n; j++) {
      if (a[i][j] == 0) {
        a[i][j] = e;
        e += 2;
      } else {
        a[i][j] = o;
        o += 2;
      }
    }
  }
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < n; j++) {
      cout << a[i][j];
      if (j < n - 1) cout << ' ';
    }
    cout << "\n";
  }
  return 0;
}