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
def run(K):
    n_max = 500
    '''
    K <= 500の場合は、n = Kとして1行につき1色を塗ればよい
        111
        222
        333
    Kが4の倍数の場合は、n = K/2以下の塗り方ができる
        1234
        5678
        3412
        7856
    上記でn+i+1色をi色に塗り替えても条件を満たす
        1234
        5178
        3412
        7851
    K = 1000の時は、n = 500で満たし、塗り替え操作で501～999まで表現できる
    '''
    n = 0
    trouts = []
    if K <= n_max:
        n = K
        for i in range(K):
            trouts.append([i+1]*K)
    else:
        n = n_max
        change = n_max*2 - K
        c1 = [i for i in range(1, n_max+1)]
        # c - 1 + n_max - c + 1 (c>0)
        # 0 + n_max (c==0)
        c2 = [i for i in range(2, change+1)] + [i for i in range(n_max+1, n_max*2+1-change)]
        if change > 0:
            c2 += [1]
        for i in range(n_max):
            shift = (i//2) * 2
            if i % 2 == 0:
                trouts.append(c1[shift:] + c1[:shift])
            else:
                trouts.append(c2[shift:] + c2[:shift])
    return (n, trouts)


def main():
    K = int(input())
    n, trouts = run(K)
    print(n)
    for i in range(len(trouts)):
        print(' '.join(map(str, trouts[i])))


if __name__ == '__main__':
    main()
