The best I could think of is 3^{log(max(a_i))}.

finding the max or is easy. consider binary representation of a number to be it’s mask.

now for a mask m let dp[m] be minimum no. of elements required such that their or is a supermask of m. so if m=6=110. then both 2,4 and 2,5 are valid choice. notice that 2,5 produces 111 but is still valid as we need a super mask.

now calculating the dp.

First remove all duplicates in a. it’s easy to see this won’t change the answer.

Initialize dp to infinity. Now for each element in a initialize dp[a$_i$] to 1 and also each of it’s submasks. Now lets say a$_i$ has k bits set in it. then it has 2$^k$ submasks.

so the entire initialization process costs \Sigma_{k=0}^{k=20} C(20,k)2$^k$. this by binomial expansion is (1+2)^{20} which is 3$^{20}$.

Now start calculating each dp[m] in incresing order of no. of bits set in m.

for each mask m consider every submask p. dp[m]=min(dp[m],dp[p]+dp[m^p]).

this relation means if p can be obtained in n1 items and m^p(which is completment of p wrt m) can be obtained from n2 items. then since their or is m, m can be obtained in n1+n2 elements.

again total no. of submasks if 2^k and therefore takes total time of 3^{20}.

final answer is dp[OR].

If any part is not clear you can ask me. Or if I’m wrong point it out. This would work easily for a$_i$<=1e5 but will tl for 1e6. maybe someone can think of some optimization.