B'Today Adilbek is taking his probability theory test. Unfortunately, when Adilbek arrived at the university, there had already been a long queue of students wanting to take the same test. Adilbek has estimated that he will be able to start the test only T seconds after coming. Fortunately, Adilbek can spend time without revising any boring theorems or formulas. He has an app on this smartphone which contains n Japanese crosswords to solve. Adilbek has decided to solve them all one by one in the order they are listed in the app, without skipping any crossword. For each crossword, a number t_i is given that represents the time it takes an average crossword expert to solve this crossword (the time is given in seconds). Adilbek is a true crossword expert, but, unfortunately, he is sometimes unlucky in choosing the way to solve the crossword. So, it takes him either t_i seconds or t_i + 1 seconds to solve the i -th crossword, equiprobably (with probability frac{1}{2} he solves the crossword in exactly t_i seconds, and with probability frac{1}{2} he has to spend an additional second to finish the crossword). All these events are independent. After T seconds pass (or after solving the last crossword, if he manages to do it in less than T seconds), Adilbek closes the app (if he finishes some crossword at the same moment, that crossword is considered solved; otherwise Adilbek does not finish solving the current crossword at all). He thinks it would be an interesting probability theory problem to calculate E -- the expected number of crosswords he will be able to solve completely. Can you calculate it? Recall that the expected value of a discrete random variable is the probability-weighted average of all possible values -- in this problem it means that the expected value of the number of solved crosswords can be calculated as E = sum limits_{i = 0}^{n} i p_i , where p_i is the p'...