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 ll long long
using namespace std;
template <typename T> void read(T &x){
	x=0;char c=getchar();int fh=1;
	while (!isdigit(c)){if (c=='-')fh=-1;c=getchar();}
	while (isdigit(c))x=x*10+c-'0',c=getchar();
	x*=fh;
}
int n,a[510][510],m;
void fu(int x,int y,int p,int q){
	for (int i=1;i<=m;i++){
		if (i%2==1)a[x][y]=p;else a[x][y]=q;
		x=(x+1)%m;y=(y+1)%m;
	}
}
int main(){
	cin>>n;
	if (n<=500){
		cout<<n<<endl;m=n;
		for (int i=1;i<=n;i++)fu(0,i-1,i,i);
	}else{
		cout<<500<<endl;m=500;
		int nu=0;
		for (int i=1;i<=n-500;i++){fu(0,i-1,nu+1,nu+2);nu+=2;}
		for (int i=n-500+1;i<=500;i++){fu(0,i-1,nu+1,nu+1);nu++;}
	}
	for (int i=0;i<m;i++){
		for (int j=0;j<m;j++)cout<<a[i][j]<<' ';cout<<endl;
	}
	return 0;
}