Let 2 be a finite alphabet and let L and L2 be two languages over 2. Assume you have access to two routines IsStringInL1(u) and IsStringinL2(u). The former routine decides whether a given string u is in L1 and the latter whether u is in L2. Using these routines as black boxes describe an efficient algorithm that given an arbitrary string w * decides whether w (LiUL2). To evaluate the running time of your solution you can assume that calls to IsStringInL10 and IsStringInL20 take constant time. Note that you are not assuming any property of L1 or L2 other than being able to test membership in those languages. Let 2 be a finite alphabet and let L and L2 be two languages over 2. Assume you have access to two routines IsStringInL1(u) and IsStringinL2(u). The former routine decides whether a given string u is in L1 and the latter whether u is in L2. Using these routines as black boxes describe an efficient algorithm that given an arbitrary string w * decides whether w (LiUL2). To evaluate the running time of your solution you can assume that calls to IsStringInL10 and IsStringInL20 take constant time. Note that you are not assuming any property of L1 or L2 other than being able to test membership in those languages