Since the sum of non-decreasing sequences is a non-decreasing sequence, and the sum of non-increasing sequences is a non-increasing sequence, then, a sumotonic sequence can be always represented by the sum of one non-decreasing sequence and one non-increasing sequence.

Moreover, since all elements in the sequence are non-negative (though they're positive as well), a sumotonic sequence guarantees

$$ \sum_{i=1}^{n-1} (a_i-a_{i+1})[a_i> a_{i+1}]\le a_1 $$

while there's no upper limit for the non-decreasing sequence but a lower limit $0$ for the non-increasing sequence.

It's stupid to use two segment trees to solve it and absolutely makes no sense. One can simply use a segment tree to maintain the reversed difference (as shown above), and use lazy tags when the current operation doesn't change the count of positive numbers in its corresponding range.

It's trivial to observe that there're at most $(n-1)$ positive reversed differences in the original sequence, and each operation introduces at most $1$ new positive reversed difference. The algorithm traverses to the bottom of the segment tree only when its positiveness changes, then the amortized time complexity is $O((n+k)\log n)$.