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;
const int MX = 5e5 + 7;
const long double eps = -0.00000001;
const long long mod = 1e9 + 7;
long long po(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long x = po(a, b / 2) % mod;
  x *= x;
  if (b % 2) {
    return x * a;
  }
  return x;
}
int dx[] = {-1, -1, 0, -1, +1, +1, +1, 0};
int dy[] = {0, -1, -1, +1, 0, -1, +1, +1};
pair<long long, long long> a[MX];
long long ans = 1e18;
long long id = -1;
int n;
int magicSquare[100][100];
void Square(int n) {
  int i = n / 2;
  int j = n - 1;
  for (int num = 1; num <= n * n;) {
    if (i == -1 && j == n) {
      j = n - 2;
      i = 0;
    } else {
      if (j == n) j = 0;
      if (i < 0) i = n - 1;
    }
    if (magicSquare[i][j]) {
      j -= 2;
      i++;
      continue;
    } else
      magicSquare[i][j] = num++;
    j++;
    i--;
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) printf("%d ", magicSquare[i][j]);
    printf("\n");
  }
}
int main() {
  cin >> n;
  memset(magicSquare, 0, sizeof(magicSquare));
  Square(n);
  return 0;
}