#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