This is the hard version of the problem. The only difference is that (x=1) in this version. You must solve both versions to be able to hack. You are given two integers (n) and (x) ((x=1)). There are (n) balls lined up in a row, numbered from (1) to (n) from left to right. Initially, there is a value (a_i) written on the (i)-th ball. For each integer (i) from (1) to (n), we define a function (f(i)) as follows: Suppose you have a set (S = \{1, 2, \ldots, i\}). In each operation, you have to select an integer (l) ((1 \leq l < i)) from (S) such that (l) is not the largest element of (S). Suppose (r) is the smallest element in (S) which is greater than (l). If (a_l > a_r), you set (a_l = a_l + a_r) and remove (r) from (S). If (a_l < a_r), you set (a_r = a_l + a_r) and remove (l) from (S). If (a_l = a_r), you choose either the integer (l) or (r) to remove from (S): If you choose to remove (l) from (S), you set (a_r = a_l + a_r) and remove (l) from (S). If you choose to remove (r) from (S), you set (a_l = a_l + a_r) and remove (r) from (S). If (a_l > a_r), you set (a_l = a_l + a_r) and remove (r) from (S). If (a_l < a_r), you set (a_r = a_l + a_r) and remove (l) from (S). If (a_l = a_r), you choose either the integer (l) or (r) to remove from (S): If you choose to remove (l) from (S), you set (a_r = a_l + a_r) and remove (l) from (S). If you choose to remove (r) from (S), you set (a_l = a_l + a_r) and remove (r) from (S). If you choose to remove (l) from (S), you set (a_r = a_l + a_r) and remove (l) from (S). If you choose to remove (r) from (S), you set (a_l = a_l + a_r) and remove (r) from (S). (f(i)) denotes the number of integers (j) ($$$1 \le j \le i$$