Problem: You've got an n × n × n cube, split into unit cubes. Your task is to number all unit cubes in this cube with positive integers from 1 to n3 so that: 

  * each number was used as a cube's number exactly once; 
  * for each 1 ≤ i < n3, unit cubes with numbers i and i + 1 were neighbouring (that is, shared a side); 
  * for each 1 ≤ i < n there were at least two different subcubes with sizes i × i × i, made from unit cubes, which are numbered with consecutive numbers. That is, there are such two numbers x and y, that the unit cubes of the first subcube are numbered by numbers x, x + 1, ..., x + i3 - 1, and the unit cubes of the second subcube are numbered by numbers y, y + 1, ..., y + i3 - 1. 



Find and print the required numeration of unit cubes of the cube.

Input

The first line contains a single integer n (1 ≤ n ≤ 50) — the size of the cube, whose unit cubes need to be numbered.

Output

Print all layers of the cube as n n × n matrices. Separate them with new lines. Print the layers in the order in which they follow in the cube. See the samples for clarifications. 

It is guaranteed that there always is a solution that meets the conditions given in the problem statement.

Examples

Input

3


Output

1 4 17 
2 3 18 
27 26 19 

8 5 16 
7 6 15 
24 25 20 

9 12 13 
10 11 14 
23 22 21 

Note

In the sample the cubes with sizes 2 × 2 × 2 are numbered with integers 1, ..., 8 and 5, ..., 12.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int mat[60][60][60][60];
void init() {
  mat[1][0][0][0] = 1;
  mat[2][0][0][0] = 1;
  mat[2][0][1][0] = 2;
  mat[2][0][1][1] = 3;
  mat[2][0][0][1] = 4;
  mat[2][1][0][1] = 5;
  mat[2][1][1][1] = 6;
  mat[2][1][1][0] = 7;
  mat[2][1][0][0] = 8;
}
void odd(int n) {
  int p = 0;
  for (int k = 0; k < n; k++) {
    if (k % 2)
      for (int j = n - 1; j >= 0; j--) mat[n][0][j][k] = ++p;
    else
      for (int j = 0; j < n; j++) mat[n][0][j][k] = ++p;
  }
  for (int i = 0; i < n - 1; i++)
    for (int j = 0; j < n - 1; j++)
      for (int k = 0; k < n - 1; k++)
        mat[n][k + 1][n - 1 - i][n - 1 - j] = mat[n - 1][i][j][k] + p;
  p += (n - 1) * (n - 1) * (n - 1);
  for (int k = n - 1; k > 0; k--) {
    if (k % 2)
      for (int i = n - 1; i > 0; i--) mat[n][i][0][k] = ++p;
    else
      for (int i = 1; i < n; i++) mat[n][i][0][k] = ++p;
  }
  for (int i = 1; i < n; i++) {
    if (i % 2)
      for (int j = 0; j < n; j++) mat[n][i][j][0] = ++p;
    else
      for (int j = n - 1; j >= 0; j--) mat[n][i][j][0] = ++p;
  }
}
void even(int n) {
  int p = 0;
  for (int k = 0; k < n; k++) {
    if (k % 2)
      for (int j = n - 1; j >= 0; j--) mat[n][0][j][k] = ++p;
    else
      for (int j = 0; j < n; j++) mat[n][0][j][k] = ++p;
  }
  for (int i = 0; i < n - 1; i++)
    for (int j = 0; j < n - 1; j++)
      for (int k = 0; k < n - 1; k++)
        mat[n][j + 1][i][n - 1 - k] = mat[n - 1][i][j][k] + p;
  p += (n - 1) * (n - 1) * (n - 1);
  for (int k = n - 1; k > 0; k--) {
    if (k % 2)
      for (int i = 1; i < n; i++) mat[n][i][n - 1][k] = ++p;
    else
      for (int i = n - 1; i > 0; i--) mat[n][i][n - 1][k] = ++p;
  }
  for (int j = n - 1; j >= 0; j--) {
    if (j % 2)
      for (int i = n - 1; i > 0; i--) mat[n][i][j][0] = ++p;
    else
      for (int i = 1; i < n; i++) mat[n][i][j][0] = ++p;
  }
}
int main() {
  init();
  for (int i = 3; i <= 50; i++) {
    if (i % 2)
      odd(i);
    else
      even(i);
  }
  int n;
  cin >> n;
  for (int k = 0; k < n; k++) {
    for (int j = 0; j < n; j++) {
      for (int i = 0; i < n; i++) cout << mat[n][i][j][k] << ' ';
      cout << endl;
    }
    cout << endl;
  }
  return 0;
}