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 PYTHON3):

Solution:
from math import sqrt, ceil
K = int(input())
 
if K <= 500:
    print(K)
    for i in range(K):
        Cout = [i+1 for n in range(K)]
        print(" ".join(map(str, Cout)))
else:
    lim = int(sqrt(K))
    for i in range(2, lim+1):
        if K % (i**2) == 0:
            print(K//i)
            for n in range(K//i):
                Cout = [i * n + (mod % i) + 1 for mod in range(K//i)]
                print(" ".join(map(str, Cout)))
            break
    else:
        n = 2 * ceil(K/4)
        print(n)
        for raw in range(n):
            if raw % 2 == 0:
                Cout = [(raw + c) % n + 1 for c in range(n)]
            else:
                Cout = [n + 1 + (raw + c) % n for c in range(n)]
                for i in range(n):
                    if Cout[i] > K: Cout[i] -= n
            print(" ".join(map(str, Cout)))