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 = = \dfrac{2^m}{3^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

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   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 shoudld be a_n. if there is any such a_i that for all 0$\leq$ j \leq$  ≠  a_i AND 2$a_i$  ≠  a_i, then this 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).