#include <iostream>
using namespace std;
int N,M;
long long int NtoT(string s){
long long ans=0;
for(int i=0;i<s.length();++i){
if(s[i]-'0'<=9&&s[i]-'0'>=0) ans=ans*N+s[i]-'0';
else ans=ans*N+s[i]-'A'+10;
}
return ans;
}
string TtoM(long long int num){
string ans;
char temp;
while(num){
if(num%M>=10){
temp=char(num%M-10+'A');
}
else {
temp=char (num%M+'0');
}
ans=temp+ans;
num/=M;
}
return ans;
}
int main()
{
int T;cin>>T;
while(T--){
cin>>N>>M;
string s;cin>>s;
string ans=TtoM(NtoT(s));
cout<<ans<<endl;
}
// 请在此输入您的代码
return 0;
}
