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
#include<bits/stdc++.h>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
using namespace std;
const int N=1005;
int n,m,a[N][N],l;
void clr(int x,int y,int v){
	for (int i=x,j=y;i<=l;i++,j--) a[i][j]=v;
}
int main(){
	scanf("%d",&n); m=((n+3)/4)*4;
	if (n==1) return printf("1\n1\n"),0;
	l=m/2; int x=1,y=l;
	rep (t,1,l){
		for (int i=x,j=y;i<=l;i++,j--) a[i][j]=2*t-(i&1);
		y++;
	}
	if (n%4!=0){
		if (n%4<=3) clr(1,2*l-1,n);
		if (n%4<=2) clr(1,2*l-2,n-1);
		if (n%4<=1) clr(1,2*l-3,n-2);
	}
	rep (i,1,l) rep (j,1,l) if (!a[i][j]) a[i][j]=a[i][j+l];
	printf("%d\n",l);
	rep (i,1,l){rep (j,1,l) printf("%d ",a[i][j]); puts("");}
	return 0;
}