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>
#define N 500
using namespace std;
int a[N+10][N+10];
int read(){
	int x=0,f=1;
	char ch=getchar();
	while (ch<'0' || ch>'9'){
		if (ch=='-') f=-1;
		ch=getchar();
	}
	while (ch>='0' && ch<='9'){
		x=x*10+ch-'0';ch=getchar();
	}
	return x*f;
}
signed main(){
	int n=read();
	if (n<=N){
		printf("%d\n",n);
		for (int i=1;i<=n;++i){
			for (int j=1;j<=n;++j){
				printf("%d ",i);
			}
			putchar('\n');
		}
	}else{
		for (int i=0;i<N;++i){
			for (int j=0;j<N;++j){
				if (i+N+1>n) a[j][(i+j)%N]=i+1;
				else a[j][(i+j)%N]=j&1?i+N+1:i+1;
			}
		}
		printf("%d\n",N);
		for (int i=0;i<N;++i){
			for (int j=0;j<N;++j){
				printf("%d ",a[i][j]);
			}
			putchar('\n');
		}
	}
	return 0;
}