A14644 | Omkar and Time Travel

El Psy Kongroo.

Omkar is watching Steins;Gate.

In Steins;Gate, Okabe Rintarou needs to complete $n$ tasks ( $1 \leq n \leq 2 \cdot 10^5$ ). Unfortunately, he doesn't know when he needs to complete the tasks.

Initially, the time is $0$ . Time travel will now happen according to the following rules:

- For each $k = 1, 2, \ldots, n$ , Okabe will realize at time $b_k$ that he was supposed to complete the $k$ -th task at time $a_k$ ( $a_k < b_k$ ).
- When he realizes this, if $k$ -th task was already completed at time $a_k$ , Okabe keeps the usual flow of time. Otherwise, he time travels to time $a_k$ then immediately completes the task.
- If Okabe time travels to time $a_k$ , all tasks completed after this time will become incomplete again. That is, for every $j$ , if $a_j>a_k$ , the $j$ -th task will become incomplete, if it was complete (if it was incomplete, nothing will change).
- Okabe has bad memory, so he can time travel to time $a_k$ only immediately after getting to time $b_k$ and learning that he was supposed to complete the $k$ -th task at time $a_k$ . That is, even if Okabe already had to perform $k$ -th task before, he wouldn't remember it before stumbling on the info about this task at time $b_k$ again.

Please refer to the notes for an example of time travelling.

There is a certain set $s$ of tasks such that the first moment that all of the tasks in $s$ are simultaneously completed (regardless of whether any other tasks are currently completed), a funny scene will take place. Omkar loves this scene and wants to know how many times Okabe will time travel before this scene takes place. Find this number modulo $10^9 + 7$ . It can be proven that eventually all $n$ tasks will be completed and so the answer always exists.

The first line contains an integer $n$ ( $1 \leq n \leq 2 \cdot 10^5$ ) — the number of tasks that Okabe needs to complete.

$n$ lines follow. The $k$ -th of these lines contain two integers $a_k$ and $b_k$ ( $1 \leq a_k < b_k \leq 2n$ ) — the time at which Okabe needs to complete the $k$ -th task and the time that he realizes this respectively. All $2n$ of these times are distinct (so every time from $1$ to $2n$ inclusive appears exactly once in the input).

The next line contains a single integer $t$ ( $1 \leq t \leq n$ ) — the size of the set $s$ of tasks that lead to the funny scene.

The last line contains $t$ integers $s_1, s_2, \ldots, s_t$ — ( $1 \leq s_k \leq n$ , the numbers $s_1, s_2, \ldots, s_t$ are distinct) — the set $s$ of tasks.

Output a single integer — the number of times that Okabe time travels until all tasks in the set $s$ are simultaneously completed, modulo $10^9 + 7$ .