After the complete simplification of the fraction $$\frac{\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)}{\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)} + \frac{\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)}{\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{5}+\frac{1}{6}\right)}$$, the result is of the form $$\frac{a}{b}$$, where a and b are natural numbers with no common factor other than 1. What fraction is to be subtracted from $$\frac{a}{b}$$ to get 1, and what is the integer part of this fraction?