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 Ans[51][51][51];
void Solve(int N) {
  if (N == 1) {
    Ans[0][0][0] = 0;
    Ans[0][1][0] = 1;
    return;
  }
  Solve(N - 1);
  static int A[51][51][51];
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++)
      for (int k = 0; k < N; k++) A[i][j][k] = Ans[i][j][k];
  int Count = 0;
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++)
      if (i & 1)
        Ans[i][0][N - 1 - j] = Count++;
      else
        Ans[i][0][j] = Count++;
  if (N & 1) {
    for (int i = 0; i < N - 1; i++)
      for (int j = 0; j < N; j++)
        for (int k = 0; k < N - 1; k++)
          Ans[N - 1 - k][i + 1][N - 1 - j] = A[i][j][k] + Count;
    Count += N * (N - 1) * (N - 1);
    for (int i = 1; i < N; i++)
      for (int j = 0; j < N; j++)
        if (i & 1)
          Ans[0][i][j] = Count++;
        else
          Ans[0][i][N - 1 - j] = Count++;
    for (int i = 0; i < N; i++) Ans[i][N][0] = Count++;
    for (int i = N - 1; i >= 0; i--)
      for (int j = 1; j < N; j++)
        if (~i & 1)
          Ans[i][N][j] = Count++;
        else
          Ans[i][N][N - j] = Count++;
  } else {
    for (int i = 0; i < N - 1; i++)
      for (int j = 0; j < N; j++)
        for (int k = 0; k < N - 1; k++)
          Ans[N - 1 - k][i + 1][j] = A[i][j][k] + Count;
    Count += N * (N - 1) * (N - 1);
    for (int i = 1; i < N; i++)
      for (int j = 0; j < N; j++)
        if (~i & 1)
          Ans[0][i][j] = Count++;
        else
          Ans[0][i][N - 1 - j] = Count++;
    for (int j = 0; j < N; j++)
      for (int i = 0; i < N; i++)
        if (~j & 1)
          Ans[i][N][j] = Count++;
        else
          Ans[N - 1 - i][N][j] = Count++;
  }
}
int main() {
  int N;
  scanf("%d", &N);
  Solve(N);
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++)
      for (int k = 0; k < N; k++) {
        printf("%d", Ans[i][j][k] + 1);
        putchar((k == N - 1) ? '\n' : ' ');
      }
    if (i + 1 < N) putchar('\n');
  }
  return 0;
}