First, note that all graphs formed by permutations are composed of a bunch of different cycles. Also, we can only create new components by disconnecting existing cycles, because nodes in the query set $S$ cannot attach to themselves or each other to create new ones. For all $1\le i\le N$, let node $i$ be pointing towards node $P_i$. Consider a node $x$ in $S$. We can delete the edge from $x$ to $P_x$ if we reroute it backwards to the node which is pointing to $x$, essentially making the edge meaningless. Deleting $D$ edges in a cycle will create $D-1$ new components. However, we can only delete an edge from $x$ to $P_x$ if the node pointing to $x$ is not in the query set, because $2$ nodes in $S$ cannot be connected with an edge. So if the node pointing to $x$ is also in $S$, then $x$ cannot disconnect the component further, and we just attach $x$ to another existing component. This does not affect the answer. Thus, we can find how many new components we can create from a cycle by counting the number of edges we can delete.

Subtask 1

The initial graph is $1$ big cycle, so our initial answer is $1$ component. Let $D$ be the number of edges we can delete. We know that deleting $D$ edges in a cycle will create $D-1$ new components. Adding $D-1$ to the initial answer $1$, we get $D$ as the final answer.

Subtask 2

Let the initial number of cycles be $C$. For each component that contains a node in the query set $S$, we can find $D$, and add $D-1$ to the answer. However, we do not actually need to consider each component independently. Notice that for each component, we always just subtract $1$. So to find the number of new components, we can just find the total $D$ value then subtract the number of components affected. Add this to the initial answer $C$ to get the final answer.