977D — Divide by three, multiply by two

1. We can see that all numbers in given sequence are distinct. since all numbers in sequence are of the form . Hence if a_i = a_j then = = 2^x - m = 3^y - n which is not possible because any power of 2 will be an even number and any power of 3 will be an odd number.

2. if there exists in sequence then 2 * a_i can not be in sequence and vice versa. We can prove it using contradiction. Suppose there is a number a_i such that both and 2 * a_i exists in sequence. by little bit of tricks this = , this again is not possible by same argument as above, we just have to change the order of exponents.

3. Hence for each a_i in sequence we see if or 2 * a_i is present(remember that only one of them can be present). Now if there is any a_i such that both and 2 * a_i is not in sequence then this should be a_n. if there is any such a_i that for all 0 ≤  j ≤ n:   ≠  a_i AND 2 * a_i  ≠  a_i, then this is a₁.

4. Once you have got a₁, you keep on producing sequence just by doing binary search for and 2 * a_i (remember only one of them is present so once you find it you print it).