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 MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int MAXN = 50;
int n, a[MAXN][MAXN], val = 1, cnt;
int main(int argc, char const *argv[]) {
  scanf("%d", &n);
  for (int i = 1; i <= n / 2 + 1; ++i) {
    for (int j = n / 2 + 1; j >= n / 2 + 1 - cnt; --j) {
      a[i][j] = val;
      val += 2;
    }
    if (cnt == 0) {
      cnt++;
      continue;
    }
    for (int j = n / 2 + 2; j <= n / 2 + 1 + cnt; ++j) {
      a[i][j] = val;
      val += 2;
    }
    cnt++;
  }
  cnt -= 2;
  for (int i = n / 2 + 2; i <= n; ++i) {
    for (int j = n / 2 + 1; j >= n / 2 + 1 - cnt; --j) {
      a[i][j] = val;
      val += 2;
    }
    if (cnt == 0) continue;
    for (int j = n / 2 + 2; j <= n / 2 + 1 + cnt; ++j) {
      a[i][j] = val;
      val += 2;
    }
    cnt--;
  }
  val = 2;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (!a[i][j]) a[i][j] = val, val += 2;
      printf("%d ", a[i][j]);
    }
    printf("\n");
  }
  return 0;
}