#include <iostream>
using namespace std;
/*248K 32MS*/
int main()
{
int s,d;
while(cin>>s>>d)
{
int count=0;
for(int i=5;i>0;i--)
{
if(s*i<(5-i)*d)
{
count=i;
break;
}
}
int total=s*count-d*(5-count);
total*=2;
if(count>=2)
total+=2*s;
else
total+=count*s-(2-count)*d;
if(total<0)
cout<<"Deficit"<<endl;
else
cout<<total<<endl;
}
}
水题 就不多说什么了，直接枚举就能ac
