Big mod function

///Parameters are string which is a huge number, divisor which can be long but can fit in long long
/// and loop initializer because string can start with negative sign
int bigmod(string num,ll divisor,int i)
{
    ll res = 0;
    for(int j=i;j<num.size();j++)
        res = (res*10+(int)num[j]-'0')%divisor;
    return res;
}