#include <iostream>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <algorithm>	// require sort next_permutation count __gcd reverse etc.
#include <cstdlib>	// require abs exit atof atoi 
#include <cstdio>		// require scanf printf
#include <functional>
#include <numeric>	// require accumulate
#include <cmath>		// require fabs
#include <climits>
#include <limits>
#include <cfloat>
#include <iomanip>	// require setw
#include <sstream>	// require stringstream 
#include <cstring>	// require memset
#include <cctype>		// require tolower, toupper
#include <fstream>	// require freopen
#include <ctime>		// require srand
#define rep(i,n) for(int i=0;i<(n);i++)
#define ALL(A) A.begin(), A.end()
/*
	No.25 有限小数

	http://yukicoder.me/problems/70


*/
using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int mod2[] = { 2, 4, 8, 6 }; 

int main()
{
	ios_base::sync_with_stdio(0);
	ll N, M; cin >> N >> M;
	ll g = __gcd (N, M );
	N /= g;
	M /= g;
	if (M != 1LL && N > M ){
		N = N - (N/M)*M;
	} // end if
	int two = 0, five = 0;
	while (M % 2LL == 0 ) M /= 2LL, two++;
	while (M % 5LL == 0 ) M /= 5LL, five++;
//	cerr << "N: " << N << " M: " << M << " g: " << g << endl;

	ll res = 1LL;
	if (two > five ){	// 2 の方が多いなら分母、分子に5の倍数を掛けて小数表示化する x/10^n
		res = 5;
	}else
	if (two < five ){
		res = mod2[(five - two - 1 + 4 ) % 4];
	} // end if

	while (N > 0 && (N % 10LL ) == 0 ) N /= 10LL;
	
	res = (res * (N % 10LL ) ) % 10LL;
	
	if (M != 1LL ){
		cout << -1 << endl;
	}else{
		cout << res << endl;
	} // end if

	return 0;
}