#include<cstdio>

using namespace std;

inline int In(){
	char c=getchar(); int x=0,ft=1;
	for(;c<'0'||c>'9';c=getchar()) if(c=='-') ft=-1;
	for(;c>='0'&&c<='9';c=getchar()) x=x*10+c-'0';
	return x*ft;
}

int main(){
	int K=In();
	if(K<=500){
		int n=K; printf("%d\n",n);
		for(int i=1;i<=n;++i) for(int j=1;j<=n;++j)
		printf("%d%c",i,j==n?'\n':' ');
	}
	if(K>500){
		int n=(K+3)/4*2; printf("%d\n",n);
		for(int i=1;i<=n;++i) for(int j=1;j<=n;++j){
			int val=(i+j)%n+(i&1?0:n)+1; if(val>K) val-=n;
			printf("%d%c",val,j==n?'\n':' ');
		}
	}
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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<cstdio>

using namespace std;

inline int In(){
	char c=getchar(); int x=0,ft=1;
	for(;c<'0'||c>'9';c=getchar()) if(c=='-') ft=-1;
	for(;c>='0'&&c<='9';c=getchar()) x=x*10+c-'0';
	return x*ft;
}

int main(){
	int K=In();
	if(K<=500){
		int n=K; printf("%d\n",n);
		for(int i=1;i<=n;++i) for(int j=1;j<=n;++j)
		printf("%d%c",i,j==n?'\n':' ');
	}
	if(K>500){
		int n=(K+3)/4*2; printf("%d\n",n);
		for(int i=1;i<=n;++i) for(int j=1;j<=n;++j){
			int val=(i+j)%n+(i&1?0:n)+1; if(val>K) val-=n;
			printf("%d%c",val,j==n?'\n':' ');
		}
	}
	return 0;
}
```