Problem:
Score : 300 points

Problem Statement

Takahashi loves the number 7 and multiples of K.

Where is the first occurrence of a multiple of K in the sequence 7,77,777,\ldots? (Also see Output and Sample Input/Output below.)

If the sequence contains no multiples of K, print -1 instead.

Constraints

1 \leq K \leq 10^6

K is an integer.

Input

Input is given from Standard Input in the following format:

K

Output

Print an integer representing the position of the first occurrence of a multiple of K. (For example, if the first occurrence is the fourth element of the sequence, print 4.)

Sample Input 1

101

Sample Output 1

4

None of 7, 77, and 777 is a multiple of 101, but 7777 is.

Sample Input 2

2

Sample Output 2

-1

All elements in the sequence are odd numbers; there are no multiples of 2.

Sample Input 3

999983

Sample Output 3

999982

Solution in Ruby (2.7.1) (judge verdict: Accepted)

require "set"
k = gets.chomp.to_i

# 7, 77, 777, ... の一般項を a_n とおくと
# k * m = a_n (mは自然数) となるような最小のnを見つけたい
# 式変形すると m = a_n / k
# 左辺は自然数なので, 右辺も自然数になる必要がある
# つまり a_n が k で割り切れるような最小のnを見つけたい
# そのようなnが存在しないときもあることに注意
# a_nは2や5の倍数ではないので, kも2や5の倍数になりえない
# 実は-1になる条件は2と5の倍数のみ

if k % 2 == 0 || k % 5 == 0
  puts -1
  exit
end

#set = Set.new
a_n = 0
n = 0
while true do
  n += 1
  a_n = (a_n * 10 + 7) % k
  break if a_n == 0
end
puts n