Problem: 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).

Examples

Input

2


Output

3
1 1 1
1 1 1
2 2 2


Input

9


Output

3
1 2 3
4 5 6
7 8 9

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#include<assert.h>
using namespace std;
int n,sz,a[505][505];
int main()
{
	scanf("%d",&n);
	if(n<=500)
	{
		printf("%d\n",n);
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
			printf("%d ",i);
			puts("");
		}
	}
	else
	{
		int k=n-500,sz=-1;
		puts("500");
		for(int i=1;i<=k;i++)
		{
			sz+=2;
			for(int j=1,k=i;j<=500;j++,k++,sz^=1)
			{
				if(j>500) j-=500;
				if(k>500) k-=500;
				a[j][k]=sz;
			}
		}
		for(int i=k+1;i<=500;i++)
		{
			++sz;
			for(int j=1,k=i;j<=500;j++,k++)
			{
				if(j>500) j-=500;
				if(k>500) k-=500;
				a[j][k]=sz;
			}
		}
		for(int i=1;i<=500;i++)
		{
			for(int j=1;j<=500;j++)
			printf("%d ",a[i][j]+1);
			puts("");
		}
	}
	return 0;
}