Familiar?

The left-to-right stack is a terrible DP state because it seems to remember a whole increasing sequence. Splitting at the minimum kills that problem. The minimum is basically a stack-clearing grenade.

Minimum Split Consider processing a contiguous block whose values are all larger than every value already in the stack. Call those earlier smaller values the external guard. They can never be popped by this block.

For any sequence $q$, let $h(q)$ be the number of its elements still present above the guard after processing it. Equivalently, these survivors are the right-to-left minima of $q$. In particular, $h(\varnothing)=0$.

Suppose the minimum value of a block occurs at position $k$. Split the block into a left part $L$, the minimum itself, and a right part $R$.

• After processing $L$, exactly $h(L)$ local elements remain.
• The minimum is smaller than every element of $L$, so it pops exactly those $h(L)$ elements. Therefore $b_k=h(L)$.
• The minimum is smaller than every element of $R$, so it is never popped.
• Processing $R$ above this minimum has exactly the same internal pop behavior as processing $R$ above any smaller guard.
• Consequently, the final height of the whole block is $1+h(R)$.

For example, in $[4,2,5,1,3]$, the stack after the left part $[4,2,5]$ contains $[2,5]$. Thus the minimum $1$ pops exactly $2$ elements. Afterward the right part $[3]$ leaves one survivor above $1$, so the whole block ends with height $1+1=2$.

This is the entire structural insight. Everything afterward is bookkeeping, albeit slightly evil bookkeeping.

Interval States The precise set of values assigned to an interval does not matter, only their relative order. For a fixed set of $r-l+1$ distinct values, define:

• $F_c(l,r)$: the number of orders on positions $[l,r]$ satisfying every specified constraint inside the interval and ending with local stack height exactly $c$.
• $G(l,r)$: the number of valid orders with arbitrary final height, so conceptually $G(l,r)=\sum_c F_c(l,r)$.

For an empty interval,

$$F_0(i,i-1)=G(i,i-1)=1,$$

and $F_c(i,i-1)=0$ for $c>0$. A nonempty interval cannot finish with height $0$, so $F_0(l,r)=0$ when $l\le r$.

Now choose position $k$ as the minimum of $[l,r]$. The constraint at $k$ checks the final height of its left part. Define

$$W(l,k)= \begin{cases} 1, & k=l\text{ and }a_k\in\{-1,0\},\\ 0, & k=l\text{ and }a_k>0,\\ G(l,k-1), & k>l\text{ and }a_k=-1,\\ F_{a_k}(l,k-1), & k>l\text{ and }a_k\ge 0. \end{cases}$$

The first two cases just say that an empty left part has height $0$.

Transitions Once the minimum is fixed at $k$, its value is forced to be the smallest value assigned to the interval. Of the remaining $r-l$ values, choose exactly $k-l$ for the left part. This contributes

$$\binom{r-l}{k-l}.$$

The remaining values go right. Standardizing either chosen set preserves all comparisons, so the interval counts apply unchanged.

If we do not care about the final height, the transition is

$$G(l,r)=\sum_{k=l}^{r} \binom{r-l}{k-l} W(l,k)G(k+1,r).$$

If the whole block must finish with height $c$, then its right part must finish with height $c-1$, because the block minimum contributes the extra surviving element. Hence, for $c\ge1$,

$$F_c(l,r)=\sum_{k=l}^{r} \binom{r-l}{k-l} W(l,k)F_{c-1}(k+1,r).$$

Each permutation has one unique minimum position and one unique set of values placed to its left, so these sums neither miss nor double-count anything.

Memory Trick Storing $F_c(l,r)$ for all $l,r,c$ takes $O(n^3)$ memory. At $n=500$, the integers alone are around 500 MB. The limit is 128 MB, so that approach is thoroughly dead.

Notice how an interval ending at $r$ is used as a left part immediately before position $r+1$:

• If $a_{r+1}=-1$, we only need $G(l,r)$.
• If $a_{r+1}=q\ge0$, we only need $F_q(l,r)$.

Computing $F_q(l,r)$ recursively requires the states $F_0,\ldots,F_q$ for intervals with the same right endpoint. Computing $G(l,r)$ requires only other $G$ states with that endpoint. Therefore, for each right endpoint $r$, store exactly one of these two families:

• one scalar $G(l,r)$ for every $l$ when $a_{r+1}=-1$;
• the $q+1$ scalars $F_0(l,r),\ldots,F_q(l,r)$ when $a_{r+1}=q$.

Append a conceptual sentinel $a_{n+1}=-1$. It makes the full interval request $G(1,n)$, which is the answer. In the code, state index $0$ is overloaded: it means $G$ for a wildcard endpoint and $F_0$ for a fixed endpoint. The endpoint tells us which meaning applies, so there is no ambiguity.

Impossible Cases Every element is pushed once and can be popped at most once. At least one element remains at the end, hence

$$\sum_{i=1}^{n} b_i\le n-1.$$

Therefore, if

$$S=\sum_{i=1}^{n}\max(a_i,0)\ge n,$$

there is no answer. Returning zero early is also what makes the ragged DP small: for every remaining test case, the sum of all requested positive heights is below $n$.

Fill Order The transition for $[l,r]$ has two dependency types:

• $W(l,k)$ uses $[l,k-1]$, which has the same left endpoint and a smaller right endpoint.
• The right factor uses $[k+1,r]$, which has the same right endpoint and a larger left endpoint.

Thus iterate $l$ from right to left, and for each $l$, iterate $r$ from left to right. Both dependency types are already computed. Empty intervals are initialized separately.

Correctness We prove that every stored state equals its definition.

For an empty interval, the only order has height $0$, so the initialization is correct.

Assume all states needed by a nonempty interval $[l,r]$ are correct. Every order on this interval has a unique minimum position $k$. By the minimum-split argument, the pop count at $k$ equals the final height of the left part, so $W(l,k)$ counts exactly the valid left orders compatible with $a_k$. The right part is processed independently above the minimum. Choosing its actual value set contributes the binomial coefficient.

For $G$, every valid right height is allowed, giving the first transition. For $F_c$, the whole interval has height $c$ exactly when the right part has height $c-1$, giving the second transition. The unique minimum position and unique left value set make all cases disjoint and exhaustive. Therefore both transitions are correct by induction.

The sentinel requests $G(1,n)$, so the reported value counts exactly all valid permutations.

Complexity For a wildcard endpoint, all transitions for that endpoint take $O(n^2)$. For a fixed next value $q$, they take $O(qn^2)$. Since the early check guarantees that the sum of all positive fixed values is below $n$, the total time is $O(n^3)$ per test case and fits the given bound on $\sum n^3$.

The ragged DP and the binomial table use $O(n^2)$ memory. All arithmetic is performed modulo $998244353$.

As a sanity check, if every $a_i=-1$, only the $G$ recurrence is used. It merely classifies every permutation by the position of its minimum, and the result is $n!$, exactly as it should be.