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 PI 3.14159265358979323846264338327950288419

using namespace std;

int k, n, subdiv, purify;
int grid[505][505];

pair<int, pair<int, int> > board_size(int x){
	for(int z = 2; z <= 500; z += 2){
		for(int subdiv = 1; subdiv <= z / 2; subdiv++){
			if(z % (2 * subdiv) != 0){
				continue;
			}
			int can = (z / 2) * (z / subdiv);
			if(can < x){
				continue;
			}
			int sub = (can - x);
			if(sub % (z / subdiv - 1) != 0){
				continue;
			}
			int purify = sub / (z / subdiv - 1);
			if(purify > z / 2){
				continue;
			}
			return {z, {subdiv, purify}};
		}
	}
	printf("WTF %d\n", x);
	assert(false);
}


int main(){
	// insurance
	for(int i = 1; i <= 1000; i++){
		board_size(i);
	}
	scanf("%d", &k);
	pair<int, pair<int, int> > res = board_size(k);
	n = res.first;
	subdiv = res.second.first;
	purify = res.second.second;
	
//	printf("%d %d %d\n", n, subdiv, purify);
	
	int cur = 0;
	for(int i = 1; i <= n / 2; i++){
		int i2 = n + 1 - i;
		if(i <= purify){
			cur++;
			for(int j = 1; j <= n; j++){
				grid[i][j] = cur;
				grid[i2][j] = cur;
			}
		}else{
			int wid = n / subdiv;
			for(int su = 0; su < subdiv; su++){
				for(int j = 1; j <= wid / 2; j++){
					grid[i][su * wid + j] = cur + j;
					grid[i2][su * wid + wid / 2 + j] = cur + j;
				}
				for(int j = 1; j <= wid / 2; j++){
					grid[i][su * wid + wid / 2 + j] = cur + wid / 2 + j;
					grid[i2][su * wid + j] = cur + wid / 2 + j;
				}
			}
			cur += wid;
		}
	}
	assert(cur == k);
	
	printf("%d\n", n);
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			printf("%d%c", grid[i][j], j == n ? '\n':' ');
		}
	}
	return 0;
}