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: Accepted)

#include <bits/stdc++.h> 
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
using namespace chrono;
#define int long long
#define ld double
#define fi first
#define se second
#define pb push_back
#define pii pair<int,int>
#define sz(x) ((long long)x.size())
#define all(x) (x).begin(), (x).end()
#define popcount(x) __builtin_popcountll(x)
const int MOD =1e9+7;

mt19937_64 rng(high_resolution_clock::now().time_since_epoch().count());
template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T> istream& operator >> (istream &in,vector<T> &A){ for(auto &x:A) in>>x; return in; };
template<typename T> ostream& operator << (ostream &out,vector<T> &A){ for(auto &x:A) out<<x<<" "; return out; };
template<typename T,typename W> istream& operator >> (istream &in,pair<T,W> &A){ in>>A.fi>>A.se; return in; };
const int N = 2*1e5+3;

void solve()
{
	int k;
	cin>>k;
	if(k%2==0)
	{
		cout<<"-1";
		return;
	}
	int ans= 0;
	int  p =1;
	for(int i=1;i<=1e8;i++)
	{
		ans=7*(p)+ans;
		ans%=k;
		if(ans==0)
		{
			cout<<i;
			return;
		}
		p*=10;
		p%=k;
	}
	cout<<"-1\n";
}
// 7, 7*11, 7*3*37 , 7*11*101,
int32_t main(){
	cin.tie(0)->sync_with_stdio(0);
	#ifdef LOCAL
	freopen("input.txt", "r", stdin);
	// freopen("output.txt", "w", stdout);
	#endif

	int t=1;
	// cin>>t;
	for(int _t=1;_t<=t;_t++){
		// cout<<"Case #"<<_t<<": ";
		solve();
	}
}