In the adjoining figure, $$\triangle ABC$$ is isosceles triangle with $$CA = CB$$. $$\angle ABC = 50^\circ$$, point $$D$$ is on side $$AB$$ such that $$\angle BCD = 46^\circ$$. Let $$\angle CAB = y$$, $$\angle ADC = x$$, $$\angle FCE = z$$, then $$x - y + 2z = $$ ______ degrees.

Since $$CA = CB$$, the base angles are equal, so $$y = \angle CAB = 50^\circ$$ and $$\angle ACB = 180^\circ - 50^\circ - 50^\circ = 80^\circ$$. In $$\triangle BCD$$, $$\angle BDC = 180^\circ - 50^\circ - 46^\circ = 84^\circ$$, so $$x = \angle ADC = 180^\circ - 84^\circ = 96^\circ$$. As $$CE$$ and $$CF$$ are the extensions of the sides $$CA$$ and $$CB$$ beyond $$C$$, the angle $$z$$ is vertically opposite $$\angle ACB$$ and equals $$80^\circ$$, giving $$x - y + 2z = 96 - 50 + 160 = 206$$.