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 maxn = 100;
int res[maxn][maxn];
int n;
bool check() {
  for (int i = 1; i <= n; ++i) {
    int sumc = 0;
    int sumr = 0;
    for (int j = 1; j <= n; ++j) {
      sumc += res[i][j];
      sumr += res[j][i];
    }
    if (sumc % 2 == 0 || sumr % 2 == 0) return false;
  }
  int sum = 0;
  for (int i = 1; i <= n; ++i) sum += res[i][i];
  if (sum % 2 == 0) return false;
  sum = 0;
  for (int i = 1; i <= n; ++i) sum += res[i][n + 1 - i];
  if (sum % 2 == 0) return false;
  return true;
}
int main() {
  scanf("%d", &n);
  int mid = n / 2 + 1;
  memset(res, 0, sizeof(res));
  res[mid][mid] = 1;
  int len = 1;
  for (int i = mid - 1; i >= 1; --i) {
    int sta = (n - len) / 2 + 1;
    for (int j = 0; j < len; ++j) res[i][sta + j] = 1;
    len += 2;
  }
  len = 1;
  for (int i = mid + 1; i <= n; ++i) {
    int sta = (n - len) / 2 + 1;
    for (int j = 0; j < len; ++j) res[i][sta + j] = 1;
    len += 2;
  }
  for (int i = 1; i <= n; ++i) res[mid][i] = 1;
  int cur0 = 2;
  int cur1 = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if (res[i][j] == 0) {
        res[i][j] = cur0;
        cur0 += 2;
      } else {
        res[i][j] = cur1;
        cur1 += 2;
      }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) printf("%d%c", res[i][j], " \n"[j == n]);
  return 0;
}