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>
using namespace std;
int g[505][505];
int main()
{
	int k,n=500;cin>>k;
	if(k <= n)
	{
		printf("%d\n", k);
		for (int i = 1; i <= k; i ++)
			for(int j = 1; j <= k; j ++)
				printf("%d%c", i, j == k ? 10 : 32);
	}
	else
	{
		puts("500");
		for (int i = 1; i <= n; i ++)
			for (int j = 1; j <= n; j ++)
			{
				int x = j, y = (i + j - 1 - 1) % n + 1;
				if (n - i + 1 == k)
				{
					g[x][y] = k;
					if (j == n) k --;
				}
				else
				{
					g[x][y] = (j & 1) ? k : k - 1;
					if (j == n) k -= 2;
				}
			}
		for (int i = 1; i <= n; i ++)
			for (int j = 1; j <= n; j ++)
				printf("%d%c", g[i][j], j == n ? 10 : 32);
	}
}