<div class="_1l1MA">if <code>target > sum(nums[i]) </code>, return <code>-1</code>. Otherwise, an answer exists</div>

<div class="_1l1MA">Solve the problem for each set bit of <code>target</code>, independently, from least significant to most significant bit. </div>

<div class="_1l1MA">For each set <code>bit</code> of <code>target</code> from least to most significant, let <code>X = sum(nums[i])</code> for <code>nums[i] <= 2^bit</code>.</div>

<div class="_1l1MA"> if <code>X >= 2^bit</code>, repeatedly select the maximum <code>nums[i]</code> such that <code>nums[i]<=2^bit</code> that has not been selected yet, until the sum of selected elements equals <code>2^bit</code>. The selected <code>nums[i]</code> will be part of the subsequence whose elements sum to target, so those elements can not be selected again. </div>

<div class="_1l1MA">Otherwise, select the smallest <code>nums[i]</code> such that <code>nums[i] > 2^bit</code>, delete <code>nums[i]</code> and add two occurences of <code>nums[i]/2</code>. Without moving to the next <code>bit</code>, go back to the step in hint 3.</div>