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 MM = 1e9 + 7;
const double eps = 1e-8;
const int MAXN = 2e3 + 10;
int n, m;
int a[MAXN][MAXN];
void read() { scanf("%d", &n); }
void solve() {
  m = n * n;
  int nn = (n + 1) / 2;
  int p = 1, q = 2;
  for (int i = 1; i <= n; i++) {
    a[nn][i] = p;
    p += 2;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
    }
  }
  for (int i = 1; i <= n; i++)
    if (i != nn) {
      a[i][nn] = p;
      p += 2;
    }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
    }
  }
  for (int i = 1; i < nn; i++)
    for (int j = 1; j < nn; j++) {
      if (p <= m) {
        a[i][j] = p;
        a[n + 1 - i][j] = p + 2;
        a[i][n + 1 - j] = p + 4;
        a[n + 1 - i][n + 1 - j] = p + 6;
        p += 8;
      } else {
        a[i][j] = q;
        a[n + 1 - i][j] = q + 2;
        a[i][n + 1 - j] = q + 4;
        a[n + 1 - i][n + 1 - j] = q + 6;
        q += 8;
      }
    }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) printf("%d%c", a[i][j], " \n"[j == n]);
  }
}
void printans() {}
int main() {
  int T = 1;
  while (T--) {
    read();
    solve();
    printans();
  }
  return 0;
}