Two bits are enough for a “hard” sum

On Jan 17 I asked the following question on

What is the complexity of the following variant of the SUBSET-SUM decision problem?

2-BIT SUBSET SUM: Given an integer $m \geq 0$, and a set of nonnegative integers $A = \{x_1, x_2, …, x_n\}$ such that every $x_i$ has at most $k=2$ bits set to $1$ ($x_i = 2^{b_{i_1}}+2^{b_{i_2}},\;\; b_{i_1},b_{i_2}\geq 0$); is there a subset $A’ \subseteq A$ such that the sum of its elements is equal to $m$ ?

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