B"Consider a sequence of digits of length 2^k [a_1, a_2, ldots, a_{2^k}] . We perform the following operation with it: replace pairs (a_{2i+1}, a_{2i+2}) with (a_{2i+1} + a_{2i+2}) bmod 10 for 0 <= i<2^{k-1} . For every i where a_{2i+1} + a_{2i+2} ge 10 we get a candy! As a result, we will get a sequence of length 2^{k-1} . Less formally, we partition sequence of length 2^k into 2^{k-1} pairs, each consisting of 2 numbers: the first pair consists of the first and second numbers, the second of the third and fourth ldots , the last pair consists of the ( 2^k-1 )-th and ( 2^k )-th numbers. For every pair such that sum of numbers in it is at least 10 , we get a candy. After that, we replace every pair of numbers with a remainder of the division of their sum by 10 (and don't change the order of the numbers). Perform this operation with a resulting array until it becomes of length 1 . Let f([a_1, a_2, ldots, a_{2^k}]) denote the number of candies we get in this process. For example: if the starting sequence is [8, 7, 3, 1, 7, 0, 9, 4] then: After the first operation the sequence becomes [(8 + 7) bmod 10, (3 + 1) bmod 10, (7 + 0) bmod 10, (9 + 4) bmod 10] = [5, 4, 7, 3] , and we get 2 candies as 8 + 7 ge 10 and 9 + 4 ge 10 . After the second operation the sequence becomes [(5 + 4) bmod 10, (7 + 3) bmod 10] = [9, 0] , and we get one more candy as 7 + 3 ge 10 . After the final operation sequence becomes [(9 + 0) bmod 10] = [9] . Therefore, f([8, 7, 3, 1, 7, 0, 9, 4]) = 3 as we got 3 candies in total. You are given a sequence of digits of length n s_1, s_2, ldots s_n . You have to answer q queries of the form (l_i, r_i) , where for i -th query you have to output f([s_{l_i}, s_{l_i+1}, ldots, s_{r_i}]) . It is guaranteed that r_i-l_i+1 is o"...