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:
#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
#define MAX 505
using namespace std;
//char nc()
//{
//	static char buf[100000],*p1=buf,*p2=buf;
//	return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
//}
char nc(){return getchar();}
int read()
{
	int x=0,y=1;
	char c=nc();
	while(!isdigit(c))
	{
		if(c=='-')y=-1;
		c=nc();
	}
	while(isdigit(c))
	{
		x=(x<<1)+(x<<3)+(c^48);
		c=nc();
	}
	return x*y;
}
int k,n;
int main()
{
	k=read();
	if(k==1) return printf("1\n1"),0;
	n=(k+1)/2;if(n&1) n++;
	printf("%d\n",n);
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<n;j++)
		{
			int now=(i+j)%n+1;
			if((i&1)&&now+n<=k) now+=n;
			printf("%d ",now);
		}
		putchar('\n');
	}
	return 0;
}