// O(dlzka odpovede) cas
#include <cstdio>
#include <set>
#include <vector>
#include <algorithm>
using namespace std;
 
int main() {
	// prejdem vsetky cisla zlozene z 5-tiek
	long long a,m,c;
	scanf(" %lld",&a);
	if(a == 5) {printf("1\n"); return 0;}
	m =5LL; // cislo zlozene z 5 modulo a
	for(int i =2; i <= 1000; i++) { // i == pocet cifier
		m =(10LL*m+5LL)%a; // pridam na koniec dalsiu 5-ku
		if(m == 0LL) {c =i; break;}} // vtedy je delitelne a
	m =5LL;
	// idem vypisat odpoved
	bool zac =true; // pozor na zaciatocne nuly
	if(m/a != 0LL) {zac =false; printf("%lld",m/a); m %=a;}
	for(int i =2; i <= c; i++) {
		m =10LL*m+5LL;
		if(m/a != 0LL && zac) zac =false;
		if(!zac) printf("%lld",m/a);
		m %=a;}
	printf("\n");
	return 0;}

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NTE=

51