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<bits/stdc++.h>

using namespace std;

#define N 1200

int n,m,c[N],f[N][N];

int main(){
	scanf("%d",&n);
	if (n==1){puts("1\n1"); return 0;}
	m=(n+3)/4*4;
	if (n%4==0){
		for (int i=1;i<=m;++i) c[i]=i;
	}
	if (n%4==1){
		c[1]=c[2]=1;
		c[3]=c[4]=2;
		c[5]=c[6]=3;
		for (int i=7;i<=m;++i) c[i]=i-3;
	}
	if (n%4==2){
		c[1]=c[2]=1;
		c[3]=c[4]=2;
		for (int i=5;i<=m;++i) c[i]=i-2;
	}
	if (n%4==3){
		c[1]=c[2]=1;
		for (int i=3;i<=m;++i) c[i]=i-1;
	}
	m/=2;
	for (int i=1;i<=m;++i)
		for (int j=1;j<=m;++j) f[j][(i+j-2)%m+1]=c[i*2-j%2];
	printf("%d\n",m);
	for (int i=1;i<=m;++i)
		for (int j=1;j<=m;++j) printf(j==m?"%d\n":"%d ",f[i][j]);
	
	return 0;
}