#include<bits/stdc++.h>
using namespace std;

int k,n,a[1010][1010];

int main()
{
	scanf("%d",&k);
	if (k==1) return puts("1"),puts("1"),0;
	for (int i=2; i<=k; i+=2) if (i*2>=k) {n=i; break;}
	for (int i=0; i<n; i+=2) 
		for (int j=0; j<n; j++) a[i][j]=(i+j)%n;
	for (int i=1; i<n; i+=2)
		for (int j=0; j<n; j++) 
		{
			a[i][j]=(i+j)%n;
			if (a[i][j]+n<k) a[i][j]+=n;
		}
	printf("%d\n",n);
	for (int i=0; i<n; i++,puts(""))
		for (int j=0; j<n; j++) printf("%d ",a[i][j]+1);
	return 0;
}
### Prompt

Generate a cpp solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

int k,n,a[1010][1010];

int main()
{
	scanf("%d",&k);
	if (k==1) return puts("1"),puts("1"),0;
	for (int i=2; i<=k; i+=2) if (i*2>=k) {n=i; break;}
	for (int i=0; i<n; i+=2) 
		for (int j=0; j<n; j++) a[i][j]=(i+j)%n;
	for (int i=1; i<n; i+=2)
		for (int j=0; j<n; j++) 
		{
			a[i][j]=(i+j)%n;
			if (a[i][j]+n<k) a[i][j]+=n;
		}
	printf("%d\n",n);
	for (int i=0; i<n; i++,puts(""))
		for (int j=0; j<n; j++) printf("%d ",a[i][j]+1);
	return 0;
}
```