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)


// Problem : C - Repsept
// Contest : AtCoder - AtCoder Beginner Contest 174
// URL : https://atcoder.jp/contests/abc174/tasks/abc174_c
// Memory Limit : 1024 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
#define vi vector<int>
#define vb vector<bool>
#define pii pair<int,int>
#define mod 1000000007
#define ss second
#define ff first
#define vpii vector<pii>
#define vvpii vector<vpii>
#define vvi vector<vi>
#define pb push_back
#define vs vector<string>
#define all(x) x.begin(),x.end()
#define mii map<int,int>
#define vd vector<double>
#define deb(x) cout<<#x<<" -> "<<x<<endl
const int inf=0x3f3f3f3f3f3f3f3f;
#define MAXN 1000005

int dx[]={0,1,1,1,0,-1,-1,-1};
int dy[]={1,1,0,-1,-1,-1,0,1};

void print(vi &v){
    for(int i=0;i<v.size();i++)cout<<v[i]<<" ";
    cout<<endl;
}

// <--------------------Code Begins Here-------------------->



int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    // cin>>t;
    while(t--){
        int n;
        cin>>n;
        int rem = 0;
	    int curr = 7 % n;
	    rem += curr;
	    rem %= n;
	    if(rem == 0) {
	    	cout<<1;
	    	return 0;
	    }
	    for(int i = 1; i < 1e7; i++) {
	    	curr = (curr * (10 % n)) % n;
	    	rem += curr;
	    	rem %= n;
		    if(rem == 0) {
		    	cout<<i + 1;
		    	return 0;
		    }
	    }
	    cout<<-1;
    }
    return 0;
}