B"The only difference between easy and hard versions is the constraints. Polycarp has to write a coursework. The coursework consists of m pages. Polycarp also has n cups of coffee. The coffee in the i -th cup Polycarp has a_i caffeine in it. Polycarp can drink some cups of coffee (each one no more than once). He can drink cups in any order. Polycarp drinks each cup instantly and completely (i.e. he cannot split any cup into several days). Surely, courseworks are not being written in a single day (in a perfect world of Berland, at least). Let's consider some day of Polycarp's work. Consider Polycarp drinks k cups of coffee during this day and caffeine dosages of cups Polycarp drink during this day are a_{i_1}, a_{i_2}, ... , a_{i_k} . Then the first cup he drinks gives him energy to write a_{i_1} pages of coursework, the second cup gives him energy to write max(0, a_{i_2} - 1) pages, the third cup gives him energy to write max(0, a_{i_3} - 2) pages, ..., the k -th cup gives him energy to write max(0, a_{i_k} - k + 1) pages. If Polycarp doesn't drink coffee during some day, he cannot write coursework at all that day. Polycarp has to finish his coursework as soon as possible (spend the minimum number of days to do it). Your task is to find out this number of days or say that it is impossible. The first line of the input contains two integers n and m ( 1 <= n <= 2 cdot 10^5 , 1 <= m <= 10^9 ) -- the number of cups of coffee and the number of pages in the coursework. The second line of the input contains n integers a_1, a_2, ... , a_n ( 1 <= a_i <= 10^9 ), where a_i is the caffeine dosage of coffee in the i -th cup. If it is impossible to write the coursework, print -1. Otherwise print the minimum number of days Polycarp needs to do it. In the first example Polycarp can drink fourth cup during first day (and write 1 page), first and secon"...