Problem:
Score : 1000 points

Problem Statement

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).

Sample Input 1

2

Sample Output 1

3
1 1 1
1 1 1
2 2 2

Every square in Color 1 has three adjacent squares in Color 1 and one adjacent square in Color 2.

Every square in Color 2 has two adjacent squares in Color 1 and two adjacent squares in Color 2.

Output such as the following will be judged incorrect:

2
1 2
2 2

3
1 1 1
1 1 1
1 1 1

Sample Input 2

9

Sample Output 2

3
1 2 3
4 5 6
7 8 9

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define maxn 1010
using namespace std;
int res[maxn][maxn];
int main()
{
	int n;
	scanf("%d", &n);
	int flag, ch1, ch2;
	for (int x = 4; x <= 500; x += 2)
	{
		for (int t1 = 0; t1 <= x / 2; t1++)
		{
			for (int t2 = 0; t2 <= x / 2; t2++)
			{
				int tmp = x * x / 4 - (x / 2 - 2) * t2 - (x / 2 - 1) * t1;
				if (tmp == n) {flag = x; ch1 = t1, ch2 = t2; break;}
			}
			if (flag) break;
		}
		if (flag) break;
	}
	int m = flag;
	printf("%d\n", m);
	int cur = 1;
	for (int i = 1; i <= ch1; i++)
	{
		for (int j = 1; j <= m; j++)
			res[i][j] = res[m - i + 1][j] = i;
		cur++;
	}
	for (int i = ch1 + 1; i <= ch1 + ch2; i++)
	{
		for (int j = 1; j <= m; j++)
			if (j % 2) res[i][j] = res[m - i + 1][j] = cur;
			else res[i][j] = res[m - i + 1][j] = cur + 1;
		cur += 2;
	}
	for (int i = ch1 + ch2 + 1; i <= m / 2; i++)
		for (int j = 1; j <= m / 2; j++)
		{
			res[i][j] = res[i][m - j + 1] = res[m - i + 1][j] = res[m - i + 1][m - j + 1] = cur;
			cur++;
		}
	for (int i = 1; i <= m; i++)
	{
		for (int j = 1; j <= m; j++)
			printf("%d ", res[i][j]);
		puts("");
	}
	return 0;
}