output_description: Print a solution in the following format:
  
      
      
      N
      a_1 a_2 ... a_N
      
  
  Here, 2 ≤ N ≤ 50 and 0 ≤ a_i ≤ 10^{16} + 1000 must hold.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      K
attempt: import sys
  import math
  import collections
  import itertools
  import array
  import inspect
  
  # Set max recursion limit
  sys.setrecursionlimit(1000000)
  
  
  # Debug output
  def chkprint(*args):
      names = {id(v): k for k, v in inspect.currentframe().f_back.f_locals.items()}
      print(", ".join(names.get(id(arg), "???") + " = " + repr(arg) for arg in args))
  
  
  # Binary converter
  def to_bin(x):
      return bin(x)[2:]
  
  
  def li_input():
      return [int(_) for _ in input().split()]
  
  
  def gcd(n, m):
      if n % m == 0:
          return m
      else:
          return gcd(m, n % m)
  
  
  def gcd_list(L):
      v = L[0]
  
      for i in range(1, len(L)):
          v = gcd(v, L[i])
  
      return v
  
  
  def lcm(n, m):
      return (n * m) // gcd(n, m)
  
  
  def lcm_list(L):
      v = L[0]
  
      for i in range(1, len(L)):
          v = lcm(v, L[i])
  
      return v
  
  
  # Width First Search (+ Distance)
  def wfs_d(D, N, K):
      """
      D: 隣接行列(距離付き)
      N: ノード数
      K: 始点ノード
      """
  
      dfk = [-1] * (N + 1)
      dfk[K] = 0
  
      cps = [(K, 0)]
      r = [False] * (N + 1)
      r[K] = True
      while len(cps) != 0:
          n_cps = []
          for cp, cd in cps:
              for i, dfcp in enumerate(D[cp]):
                  if dfcp != -1 and not r[i]:
                      dfk[i] = cd + dfcp
                      n_cps.append((i, cd + dfcp))
                      r[i] = True
  
          cps = n_cps[:]
  
      return dfk
  
  
  # Depth First Search (+Distance)
  def dfs_d(v, pre, dist):
      """
      v:  現在のノード
      pre: １つ前のノード
      dist: 現在の距離
  
      以下は別途用意する
      D: 隣接リスト(行列ではない)
      D_dfs_d: dfs_d関数で用いる，始点ノードから見た距離リスト
      """
  
      global D
      global D_dfs_d
  
      D_dfs_d[v] = dist
  
      for next_v, d in D[v]:
          if next_v != pre:
              dfs_d(next_v, v, dist + d)
  
      return
  
  
  # --------------------------------------------
  
  dp = None
  
  
  def main():
      e = 0.00001
      K = int(input())
  
      if K % 2 == 1:
          print(2)
          print(int((K * 0.5) + 2.5 + e), int((K * 0.5) - 0.5 + e))
      else:
          print(2)
          print(int((K * 0.5) + 1 + e), int((K * 0.5) + 1 + e))
  
  
  main()
  
problem_description: Statement
  
  We have a sequence of length N consisting of non-negative integers. Consider
  performing the following operation on this sequence until the largest element
  in this sequence becomes N-1 or smaller.
  
    * Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.
  
  It can be proved that the largest element in the sequence becomes N-1 or
  smaller after a finite number of operations.
  
  You are given an integer K. Find an integer sequence a_i such that the number
  of times we will perform the above operation is exactly K. It can be shown
  that there is always such a sequence under the constraints on input and output
  in this problem.
samples: [{"input": "0", "output": "4\n    3 3 3 3\n    \n\n* * *"}, {"input": "1", "output": "3\n    1 0 3\n    \n\n* * *"}, {"input": "2", "output": "2\n    2 2\n    \n\nThe operation will be performed twice: [2, 2] -> [0, 3] -> [1, 1].\n\n* * *"}, {"input": "3", "output": "7\n    27 0 0 0 0 0 0\n    \n\n* * *"}, {"input": "1234567894848", "output": "10\n    1000 193 256 777 0 1 1192 1234567891011 48 425"}]