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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <sstream>
#include <vector>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <utility>
#include <map>
#include <list>
#include <climits>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <bitset>
#include <string>

// #define cerr if(false)cerr
#define watch(x) cerr << "> " << #x << ": " << x << "\n";

using namespace std;

template <typename T>
std::ostream &operator <<(std::ostream &out, vector<T> &v) {
	for (typename vector<T>::size_type i = 0; i < v.size(); ++i)
		out << v[i] << " ";
	out << "\n";
    return out;
}
template <typename T, typename N>
std::ostream &operator <<(std::ostream &out, vector<pair<T, N> > &v) {
	for (size_t i = 0; i < v.size(); ++i)
		out << "(" << v[i].first << ", " << v[i].second << ") ";
	out << "\n";
    return out;
}
template <typename T>
std::ostream &operator <<(std::ostream &out, vector<vector<T> > &v) {
	for (size_t i = 0; i < v.size(); ++i) {
		for (size_t j = 0; j < v[i].size(); ++j) {
			out << v[i][j] << " ";
		}
		out << "\n";
	}
   	return out;
}

int main() {

	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	int k;
	cin >> k;

	if (k % 2 == 0) {
		cout << "-1\n";
	} else {
		int s = 7;
		int ans = 1;
		while (ans < k) {
			if (s % k == 0) {
				cout << ans << "\n";
				return 0;
			}

			s = (s % k) * 10 + 7;
			++ans;
		}

		cout << "-1\n";
	}

	return 0;

}