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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
using namespace std;
 
#define mod 1000000007
#define FOR(x,to) for(int x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define long long long
inline int rei(){int x;cin>>x;return x;}
inline long rel(){long x;cin>>x;return x;}
inline string res(){string x;cin>>x;return x;}
//------------------------------------------------------- 
void Calc(){
	int K = rei();
	if(K == 1){
		cout << 1 << endl;
		cout << 1 << endl;
	}
	int N = (K+1)/2;
	if(N % 2){
		N++;
	}
	cout << N << endl;
	for(int i=0;i<N;i++){
		if(i % 2 == 0){
			for(int j=0;j<N;j++){
				cout << (i+j) % N + 1;
				if(j != N-1){
					cout << " ";
				}
				else{
					cout << endl;
				}
			}
		}	
		else{
			for(int j=0;j<N;j++){
				int c = (i+j) % N;
				if(c+N < K){
					c += N;
				}
				cout << c + 1;
				if(j != N-1){
					cout << " ";
				}
				else{
					cout << endl;
				}
			}
		}
	}
}
int main(int argc,char** argv){
	ios::sync_with_stdio(false), cin.tie(0);
	cout.tie(0); Calc(); return 0;
}