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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define pb push_back
#define pll pair <ll, ll>
#define mp make_pair
#define pyshnapyshnakaa ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
// #define x first
// #define y second
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#define plll pair <pair <ll, ll>, ll>
#define pllll pair <pair <ll, ll>, pair <ll, ll> >
#define psl pair <string, ll>
#define pld pair <ld, ld>
#define all(a) a.begin(), a.end()
#define vvl vector <vector <ll> >
#define cld complex <double>

typedef long long ll;

typedef long double ld;

using namespace std;

const ll maxn = 1e6 + 100;

ll n, m, k, t;

int main() {
    ll q, w, e, a, b, c;
    cin >> k;
    ll half = (k + 3) / 4;
    n = half * 2;
    cout << n << endl;
    for (q = 0; q < n; q++) {
        for (w = 0; w < n; w++) {
            if (q % 2 == 0) {
                cout << (q + w) % n + 1 << " ";
            }
            else {
                if ((q + w) % n + n < k) {
                    cout << (q + w) % n + n + 1 << " ";
                } 
                else {
                    cout << (q + w) % n + 1 << " ";
                }
            }
        }
        cout << endl;
    }
    return 0;
}