human
What is a the Python program you requested given the following:
Even if the world is full of counterfeits, I still regard it as wonderful.

Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this.

The phoenix has a rather long lifespan, and reincarnates itself once every a! years. Here a! denotes the factorial of integer a, that is, a! = 1 × 2 × ... × a. Specifically, 0! = 1.

Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of b! years, that is, . Note that when b ≥ a this value is always integer.

As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge.

Input

The first and only line of input contains two space-separated integers a and b (0 ≤ a ≤ b ≤ 1018).

Output

Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.

Examples

Input

2 4

Output

2

Input

0 10

Output

0

Input

107 109

Output

2

Note

In the first example, the last digit of  is 2;

In the second example, the last digit of  is 0;

In the third example, the last digit of  is 2.
Solve the task in CPP.
assistant
#Here is the Python program you requested.

#include <bits/stdc++.h>
using namespace std;
using lli = long long int;
using pll = pair<lli, lli>;
using vi = vector<lli>;
using vvi = vector<vi>;
using pii = tuple<lli, lli>;
using vii = vector<pii>;
using vvii = vector<vii>;
string TYPE(const int*) { return "%d"; }
string TYPE(const lli*) { return "%lld"; }
string TYPE(const char*) { return "%c"; }
string TYPE(const char**) { return "%s"; }
string TYPE(const unsigned int*) { return "%u"; }
const int MAX_BUF = 100 * 1000 + 42;
char buf[MAX_BUF];
void RD() {}
template <typename T, typename... Args>
void RD(T* v, Args... args) {
  scanf((" " + TYPE(v)).c_str(), v);
  RD(args...);
}
template <typename... Args>
void RD(string* v, Args... args) {
  scanf(" %s", buf);
  (*v) = buf;
  RD(args...);
}
void PR(bool nl = true) {
  if (nl) printf("\n");
}
template <typename T, typename... Args>
void PR(bool nl, T v, Args... args) {
  printf((TYPE(&v) + " ").c_str(), v);
  PR(nl, args...);
}
template <typename... Args>
void PR(bool nl, string& v, Args... args) {
  printf("%s", v.c_str());
  PR(nl, args...);
}
template <typename... Args>
void PR(Args... args) {
  PR(true, args...);
}
const long long int oo = 1000 * 1000 * 1000;
struct Coord {
  int x, y;
  Coord(int x = 0, int y = 0) : x(x), y(y) {}
  Coord operator+(const Coord& droite) const {
    return Coord(x + droite.x, y + droite.y);
  }
};
struct AB {
  int k;
  vector<lli> arbre;
  AB(int _k = 20, lli def = 0) {
    k = _k;
    for (lli i = 0; i < (lli)(1 << k); ++i)
      arbre.push_back(i < (1 << (k - 1)) ? 0LL : def);
    for (lli i = (((1 << (k - 1)) - 1)); i >= (lli)(1); --i)
      arbre[i] = arbre[i << 1] + arbre[(i << 1) ^ 1];
  }
  void set(int i, lli x) {
    int feuille = i + (1 << (k - 1));
    arbre[feuille] = x;
    iset(feuille >> 1);
  }
  void iset(int noeud) {
    if (noeud) {
      arbre[noeud] = arbre[noeud << 1] + arbre[(noeud << 1) ^ 1];
      iset(noeud >> 1);
    }
  }
  lli sum(int deb, int fin, int noeud = 1, int p = 0, int q = -1) {
    if (q < p) q = 1 << (k - 1);
    if (deb <= p &&...