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: Wrong Answer)

# -*- coding: utf-8 -*-
"""
Created on Sun Aug  2 21:35:25 2020

@author: aruto
"""

import math

def make_divisors(n):
    lower_divisors , upper_divisors = [], []
    i = 1
    while i*i <= n:
        if n % i == 0:
            lower_divisors.append(i)
            if i != n // i:
                upper_divisors.append(n//i)
        i += 1
    return lower_divisors + upper_divisors[::-1]

def power_func(a,n,p):
        bi = str(format(n,"b"))#2進表現に
        res = 1
        for i in range(len(bi)):
            res = (res*res) %p
            if bi[i] == "1":
                res = (res*a) %p
        return res
    
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

N = int(input())
if N%7 == 0:
    N = N//7

if N%5 == 0 or N%2 == 0:
    print(-1)

elif N == 1:
    print(1)

else:
    a = 1
    i = 1
    
    A = factorization(N)
    
    g = 1
    
    for n in A:
        if n[0] == 3:
            k_f = 3
        else:
            N = n[0]*9
            K = make_divisors(n[0]-1)
            for k in reversed(K):
                a = power_func(10,k,N)-1
                if a == 0:
                    k_f = k
        g = g*k_f*n[0]**(n[1]-1)//math.gcd(g,k_f*n[0]**(n[1]-1))
    print(g)