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 Python (3.8.2) (judge verdict: Time Limit Exceeded)

k = int(input())

def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])

    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])

    return arr

pn=factorization(k)

go =0
keta = len(str(k))
sevens = []

if k % 2 == 0 or k %5 == 0:
    print(-1)
elif str(k)[-1] == "3":
    for i in range(9,10**100000,10):
        if not str(k * i).count('7') == len(str(k*i)):
            continue
        else:
            print(str(k * i).count('7'))
elif str(k)[-1] == "7":
    for i in range(1,10**100000,10):
        if not str(k * i).count('7') == len(str(k*i)):
            continue
        else:
            print(str(k * i).count('7'))
elif str(k)[-1] == "9":
    for i in range(3,10**100000,10):
        if not str(k * i).count('7') == len(str(k*i)):
            continue
        else:
            print(str(k * i).count('7'))
elif str(k)[-1] == "1":
    for i in range(7,10**100000,10):
        if not str(k * i).count('7') == len(str(k*i)):
            continue
        else:
            print(str(k * i).count('7'))