// includes
#include <cstdio>
#include <cstdint>
#include <iostream>
#include <iomanip>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <algorithm>
#include <utility>
#include <functional>
#include <cmath>
#include <climits>
#include <bitset>
#include <list>
#include <random>

// macros
#define ll long long int
#define pb emplace_back
#define mk make_pair
#define pq priority_queue
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())

using namespace std;

//  types
typedef pair<int, int> P;
typedef pair<ll, int> Pl;
typedef pair<ll, ll> Pll;
typedef pair<double, double> Pd;
 
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1e9 + 7;

// solve
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}

template <typename T>
T power(T a, T n, T mod) {
  T res = 1;
  T tmp = n;
  T curr = a;
  while(tmp){
    if(tmp % 2 == 1){
      res = (T)((ll)res * curr % mod);
    }
    curr = (T)((ll)curr * curr % mod);
    tmp >>= 1;
  }
  return res;
}

int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  string n, k;
  cin >> n >> k;
  int curr = 1;
  int res = 0;
  reverse(all(n));
  for(int i = 0; i < sz(n); i++){
    int num = int(n[i] - '0');
    res = (res + num * curr % 6) % 6;
    curr = curr * 10 % 6;
  }
  if(res == 0)res = 6;
  int p = 0;
  int tmp = res;
  do{
    tmp = tmp * res % 6;
    if(tmp == 0)tmp = 6;
    p++;
  }while(tmp != res);

  curr = 1;
  int ans = 0;
  reverse(all(k));
  for(int i = 0; i < sz(k); i++){
    int num = int(k[i] - '0');
    ans = (ans + num * curr % p) % p;
    curr = curr * 10 % p;
  }
  if(ans == 0)ans += p;
  int cnt = power(res, ans, 6);
  int d[6] = {2, 8, 5, 7, 1, 4};
  cout << d[(cnt+5)%6] << endl;
	return 0;
}