B'The only difference between the easy and the hard version is the limit to the number of queries. This is an interactive problem. There is an array a of n different numbers. In one query you can ask the position of the second maximum element in a subsegment a[l..r] . Find the position of the maximum element in the array in no more than 40 queries. A subsegment a[l..r] is all the elements a_l, a_{l + 1}, ..., a_r . After asking this subsegment you will be given the position of the second maximum from this subsegment in the whole array. The first line contains a single integer n (2 <= q n <= q 10^5) -- the number of elements in the array. You can ask queries by printing "? l r " (1 <= q l < r <= q n) . The answer is the index of the second maximum of all elements a_l, a_{l + 1}, ldots, a_r . Array a is fixed beforehand and can 't be changed in time of interaction. You can output the answer by printing "! p ", where p is the index of the maximum element in the array. You can ask no more than 40 queries. Printing the answer doesn 't count as a query. After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use: Hacks To make a hack, use the following test format. In the first line output a single integer n (2 <= q n <= q 10^5) . In the second line output a permutation of n integers 1 to n . The position of n in the permutation is the position of the maximum In the sample suppose a is [5, 1, 4, 2, 3] . So after asking the [1..5] subsegment 4 is second to max value, and it 's position is 3 . After asking the [4..5] subsegment 2 is second to max value and it 's position in the whole array is 4 . Note that there are other arrays a that would produce the same interaction, and the answer for them might be different. Example o'...