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: Accepted)

#include <bits/stdc++.h>
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define SORT(a) sort(a.begin(), a.end())
#define REVERSE(a) reverse(a.begin(), a.end())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define int long long
#define INF 1000000000000000000
using namespace std;

#define ANS(f) if(f) cout << "YES" << endl; else cout << "NO" << endl;

typedef vector<int> vec;
typedef vector<vec> mat;
typedef pair<int, int> Pii;

template<typename T>
void readv(vector<T> &a){ REP(i, a.size()) cin >> a[i]; }
void readi(vector<int> &a){ REP(i, a.size()){cin >> a[i]; a[i]--;} }
void debug(mat m){REP(i, m.size()){ REP(j, m[i].size()){ cout << m[i][j] << ","; } cout << endl; }}



signed main(){

    int K; cin >> K;
    mat c;
    int n;
    if(K <= 2){
        n = K;
        c = mat(n, vec(n));
        REP(i, n) REP(j, n) c[i][j] = i;
    }else{
        n = 2;
        while(2 * n < K) n += 2;
        c = mat(n, vec(n));
        REP(i, n) REP(j, n){
            c[i][j] = n * (i % 2) + (i + j) % n;
            if(c[i][j] >= K) c[i][j] -= n;
        }
    }

    cout << n << endl;
    REP(i, n){
        REP(j, n) cout << c[i][j] + 1 << " ";
        cout << endl;
    }
    
    return 0;
}