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<stdio.h>
#include<stdlib.h>
#include<string.h>

#define ll long long
#define INF 1<<31-1
#define LNF 9223372036854775807
#define PI 3.14159265358979
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))
#define FOR(i,a,n) for(i=a;i<n;i++)
#define MOD 1000000007
//#define MOD 998244353
#define NUM 200010
 
void swap(int* a,int* b){int tmp=*a;*a=*b;*b=tmp;}
int abs(int a){if(a<0)return -a;return a;}
int dup(int a,int d){if(a%d==0)return a/d;else return a/d+1;}
int mod(int a,int m){if(a<0)return a%m+m;else return a%m;}

void swapl(ll* a,ll* b){ll tmp=*a;*a=*b;*b=tmp;}
ll absl(ll a){if(a<0)return -a;return a;}
ll dupl(ll a,ll d){if(a%d==0)return a/d;else return a/d+1;}
ll modl(ll a,ll m){if(a<0)return a%m+m;else return a%m;}

ll GCD(ll a,ll b){ll c=1,tmp=MAX(a,b);b=MIN(a,b);a=tmp;while(c!=0){c=a%b;a=b;b=c;}return a;}
ll LCM(ll a,ll b){return a*b/GCD(a,b);}

ll factorial(ll n, ll m) {return n < m ? 1 : n*factorial(n-1, m);}

int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}

// ll dp[]

int main(){
  ll k;
  ll x=0;
  ll ans = -1;

  scanf("%lld", &k);

  if(k%2 == 0){
      printf("-1\n");
      return 0;
  }
  if(k%7 == 0){
      k = k%7;
  }

  for(int i=1; i<=k; i++){
      x = (10*x+1) % k;
      if(x == 0){
          ans = i;
          break;
      }
  }
  printf("%lld\n", ans);
  return 0;
}