F. Crossword Expert
combinatorics; dp; number theory; probabilities; two pointers
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 probability that Adilbek will solve exactly \(i\) crosswords. We can represent \(E\) as rational fraction \(\frac{P}{Q}\) with \(Q > 0\). To give the answer, you should print \(P \cdot Q^{-1} \bmod (10^9 + 7)\).
The first line contains two integers \(n\) and \(T\) (\(1 \le n \le 2 \cdot 10^5\), \(1 \le T \le 2 \cdot 10^{14}\)) — the number of crosswords and the time Adilbek has to spend, respectively.The second line contains \(n\) integers \(t_1, t_2, \dots, t_n\) (\(1 \le t_i \le 10^9\)), where \(t_i\) is the time it takes a crossword expert to solve the \(i\)-th crossword.Note that Adilbek solves the crosswords in the order they are given in the input without skipping any of them.
Print one integer — the expected value of the number of crosswords Adilbek solves in \(T\) seconds, expressed in the form of \(P \cdot Q^{-1} \bmod (10^9 + 7)\).
The answer for the first sample is equal to \(\frac{14}{8}\).The answer for the second sample is equal to \(\frac{17}{8}\).