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 <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int N=5e2+5;
int n,k,tot,nxt[N],mat[N][N];

int main() {
	scanf("%d",&k),printf("%d\n",min(k,500));
	if(k<=500) for(int i=1;i<=k;++i,puts("")) for(int j=1;j<=k;++j) printf("%d ",i);
	else {
		n=500,k-=n,tot=0;
		for(int i=1;i<n;++i) nxt[i]=i+1;nxt[n]=1;
		for(int i=1;i<=n;++i,k-=(bool)(k)) {
			int a=++tot,b=k?++tot:tot;
			for(int x=1,y=i;x<=n;++x,y=nxt[y]) mat[x][y]=(x&1)?a:b;
		}
		for(int i=1;i<=n;++i,puts("")) for(int j=1;j<=n;++j) printf("%d ",mat[i][j]);
	}
	return 0;
}