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
K  = int(input())
n = 0
n1 = 0
n2 = 0
n3 = 0
n4 = 0
n5 = 0


#Kが500以下の場合、斜め作戦にする
if K <= 499:
   n = K
   mat = [[0 for n1 in range(K)] for n2 in range(K)]
   for n1 in range(K):
      for n2 in range(K-n1):
         mat[n1][n1+n2] = n2+1
   for n1 in range(K):
      for n2 in range(n1):
         mat[n1][n2] = K+n2-n1+1

#Kが500以上の場合、斜めに1マスずつしのびこませる   
else:
   n = 500
   mat = [[0 for n1 in range(500)] for n2 in range(500)]
   for n1 in range(500):
      for n2 in range(500-n1):
         mat[n1][n1+n2] = n2+1
   for n1 in range(500):
      for n2 in range(n1):
         mat[n1][n2] = 500+n2-n1+1

   #Kが500以上の場合のギザギザ部分
   for n2 in range(K-500):
      for n1 in range(250):
          kari = n1*2+n2
          if kari >= 500:
             kari = kari-500
          mat[n1*2][kari] = n2+501


#行列の出力かんがえとく 
print(n)

if K <= 500:
   out = ""
   for n1 in range(K):
      out = ""   
      for n2 in range(K):
         out = out + " " + str(mat[n1][n2])
      print(out)
   
#Kが500以上の場合、斜めに1マスずつしのびこませる   
else:
   out = ""
   for n1 in range(500):
       out = ""   
       for n2 in range(500):
          out = out + " " + str(mat[n1][n2])
       print(out)



