Problem: 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).

Examples

Input

2


Output

3
1 1 1
1 1 1
2 2 2


Input

9


Output

3
1 2 3
4 5 6
7 8 9

Solution (in JAVA):

Solution:
    import java.util.*;
    public class Main {
    	public static void main(String[] args){
	    Scanner sc = new Scanner(System.in);
	    // 整数の入力
	    int k = sc.nextInt();
	    // 出力

	    if(k == 1){
		System.out.println("1");
		System.out.println("1");
	    }else if(k == 2){
		System.out.println("2");
		System.out.println("1 2");
		System.out.println("2 1");
	    }else if(k == 3){
		System.out.println("2");
		System.out.println("1 2");
		System.out.println("3 1");
	    }else if(k == 4){
		System.out.println("2");
		System.out.println("1 2");
		System.out.println("3 4");
	    }else{
		int cut = 4 - k % 4;
		int n;
		int amari;
		if (cut != 4){
		    n = (k + cut) / 2;
		}else{
		    n = k / 2;
		    cut = 0;
		}
		System.out.println(n);
		for(int r = 0; r < n; r++){
		    for(int c = 0; c < n; c++){
			if (r % 2 == 0){
			    System.out.print((c + r) % n + 1 + " ");
			}else{
			    amari = (c + r) % n;
			    if(amari == n-1 && cut >= 1 ||
			       amari == n-2 && cut >= 2 ||
			       amari == n-3 && cut >= 3){
				System.out.print((c + r) % n + 1 + " ");
			    }else{
				System.out.print((c + r) % n + n + 1 + " ");
			    }
			}
		    }
		    System.out.println();
		}
	    }
	}
    }