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 <iostream>
#include <algorithm>
#include <vector>
#define fs first
#define se second

using namespace std;
typedef long long llong;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<llong, llong> pll;
typedef pair<ld, ld> pdd;

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    int k;
    cin >> k;
    if (k <= 500) {
        printf("%d\n", k);
        for (int i = 1; i <= k; ++i) {
            for (int j = 1; j <= k; ++j) {
                printf("%d%c", i, "\n "[j < k]);
            }
        }
        return 0;
    }
    int n = ((k + 1) / 2 + 1) / 2 * 2;
    int col[501][2] = {};
    for (int i = 1; i <= n; ++i) {
        col[i][0] = i;
    }
    for (int i = n + 1; i <= k; ++i) {
        col[i - n][1] = i;
    }
    printf("%d\n", n);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            int C = (i + j) % n + 1;
            int p = col[C][0];
            if ((j & 1) && col[C][1]) p = col[C][1];
            printf("%d%c", p, "\n "[j < n]);
        }
    }
    return 0;
}