Problem:
Score : 1000 points

Problem Statement

For an n \times n grid, let (r, c) denote the square at the (r+1)-th row from the top and the (c+1)-th column from the left.
A good coloring of this grid using K colors is a coloring that satisfies the following:

Each square is painted in one of the K colors.

Each of the K colors is used for some squares.

Let us number the K colors 1, 2, ..., K. For any colors i and j (1 \leq i \leq K, 1 \leq j \leq K), every square in Color i has the same number of adjacent squares in Color j. Here, the squares adjacent to square (r, c) are ((r-1)\; mod\; n, c), ((r+1)\; mod\; n, c), (r, (c-1)\; mod\; n) and (r, (c+1)\; mod\; n) (if the same square appears multiple times among these four, the square is counted that number of times).

Given K, choose n between 1 and 500 (inclusive) freely and construct a good coloring of an n \times n grid using K colors.
It can be proved that this is always possible under the constraints of this problem,

Constraints

1 \leq K \leq 1000

Input

Input is given from Standard Input in the following format:

K

Output

Output should be in the following format:

n
c_{0,0} c_{0,1} ... c_{0,n-1}
c_{1,0} c_{1,1} ... c_{1,n-1}
:
c_{n-1,0} c_{n-1,1} ... c_{n-1,n-1}

n should represent the size of the grid, and 1 \leq n \leq 500 must hold.
c_{r,c} should be an integer such that 1 \leq c_{r,c} \leq K and represent the color for the square (r, c).

Sample Input 1

2

Sample Output 1

3
1 1 1
1 1 1
2 2 2

Every square in Color 1 has three adjacent squares in Color 1 and one adjacent square in Color 2.

Every square in Color 2 has two adjacent squares in Color 1 and two adjacent squares in Color 2.

Output such as the following will be judged incorrect:

2
1 2
2 2

3
1 1 1
1 1 1
1 1 1

Sample Input 2

9

Sample Output 2

3
1 2 3
4 5 6
7 8 9

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main(){
  int i,j,k,n,m,d[510][510];
  /*for(i=1;i<=1000;i++){
    for(j=2;j<1000;j++){
      if(i%j==0)goto NEXT;
    }
    for(j=3;j<=500;j+=2){
      if((j*j+1)%i==0)goto NEXT;
    }
    printf("%d\n",i);
  NEXT:;
  }//*/
  scanf("%d",&n);
  for(i=2;i<=500;i++){
    if(i%n==0){
      printf("%d\n",i);
      for(j=k=0;j<i*i;j++){
	printf("%d%c",k+1,j%i==i-1?'\n':' ');
	k=(k+1)%n;
      }
      return 0;
    }	
  }
  for(i=3;i<=500;i+=2){
    if((i*i+1)%n==0)break;
  }
  m=i;
  for(i=j=0;2*i<m*m;i++){
    d[i/m][i%m]=d[m-i/m-1][m-i%m-1]=j+1;
    j=(j+1)%n;
  }
  printf("%d\n",m);
  for(i=0;i<m;i++){
    for(j=0;j<m-1;j++)printf("%d ",d[i][j]);
    printf("%d\n",d[i][m-1]);
  }
  return 0;
 
}