#include<bits/stdc++.h>
using namespace std;
#define FR0(i,N) for(i=0;i<(N);i++)
#define FR1(i,N) for(i=1;i<=(N);i++)
#define FRN(i,k,N) for(i=k;i<(N);i++)
#define pf printf
#define db double
#define max3(a,b,c) max(max(a,b),c)
#define min3(a,b,c) min(min(a,b),c)
#define sci(n) scanf("%d",&n)
#define scl(n) scanf("%lld",&n)
#define scf(n) scanf("%f",&n)
#define scd(n) scanf("%lf",&n)
#define scs(s) scanf("%s",&s)
#define scll(n) scanf("%%I64d",&n)
#define PI acos(-1.0)
#define LL long long
#define MX 1000005
#define MOD 1000000007
typedef long long int ll;
bool status[1100002];
/* Problem description:
If any problem need find last digit of m^n .Then you find base one digit and power last two digit using;
*/
int mypow(int base,int power)
{
int total;
if(base==0) return 0;
if(power==0)
{
total=1;
}
else
{
total=base;
}
for(int i=1;i<power;i++)
{
total=total*base;
total=total%10;
}
return total;
}
int main()
{
int ibase,ipower,temp;
string sbase,spower;
while(cin>>sbase>>spower)
{
if(sbase=="0" && spower=="0") break;
ibase=sbase[sbase.size()-1]-48;
if(spower.size()>=2)// we need n's last two digit
{
temp=spower[spower.size()-1]-48;
ipower=(spower[spower.size()-2]-48)*10+temp;
}
else
{
ipower=spower[spower.size()-1]-48;
}
if(ipower==0)//last two digit zero then
{
temp=mypow(ibase,99);//_m ^ 100 = _m ^ 99 * _m
temp=(temp*ibase)%10;
}
else
{
temp=mypow(ibase,ipower)%10;
}
cout<<temp<<endl;
}
}
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