Problem:
Score : 600 points

Problem 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.

Constraints

0 ≤ K ≤ 50 \times 10^{16}

Input

Input is given from Standard Input in the following format:

K

Output

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.

Sample Input 1

0

Sample Output 1

4
3 3 3 3

Sample Input 2

1

Sample Output 2

3
1 0 3

Sample Input 3

2

Sample Output 3

2
2 2

The operation will be performed twice: [2, 2] -> [0, 3] -> [1, 1].

Sample Input 4

3

Sample Output 4

7
27 0 0 0 0 0 0

Sample Input 5

1234567894848

Sample Output 5

10
1000 193 256 777 0 1 1192 1234567891011 48 425

Solution in Python (3.4.3) (judge verdict: Accepted)

#coding=UTF-8

# 基本的に N=1 seq=[K] であれば簡単だった
# 全体では毎回 N減って N-1増えるのできっと収束する
# もし、2本立てると

# 7 7
# 5 8
# 6 6
# 4 7
# 5 5
# 3 6
# 4 4
# 2 5
# 3 3
# 1 4
# 2 2
# 0 3
# 1 1 done! の12回

# f(0,0)=0
# f(1,1)=0
# f(2,2): -> 0,3 -> 1,1 =2
# f(3,3): -> 1,4 -> 2,2 + f(2,2) = 4
# f(4,4): -> 2,5 -> 3,3 + f(3,3) = 6
# f(5,5): -> 3,6 -> 4,4 + f(4,4) = 8 ...

# f(A,A)=(A-1)*2 (A>=1)

# f(2,2,2)=0
# f(3,3,3): -> 0,4,4 -> 1,1,5 -> 2,2,2 =3
# K=1のときは
#  3,0 -> 1,1 ok (非負)

# N=50にするのがつよそう
# 最大の余りは49なので
# 49個が1回ずつ殴られるパターン
# L+50-48 L+50-48 ... L+50-48 L-49 が49回後には
# L L L ... となる
# これが合法手になるのはL>=49のときである

# f([49]*50)=0
# f([50]*50)= 50 + f([49]*50)=50
# f([L]*50)=50*(L-49) (for L >= 49)

# K=50*(10**16)-1 では
# L-49=10**16-1
# L=10**16+48
# L+50-48=10**16 + 50 ok
K=int(input())
print(50)
L=K//50+49
amari=K%50
ansstr=''
#extra amari times
for idx in range(0,50,1):
    if idx < amari:
        ansstr+=str(L+50-(amari-1))+' '
    else:
        ansstr+=str(L-amari)+' '

print(ansstr[0:len(ansstr)-1])