description: El Psy Kongroo.
  
  Omkar is watching Steins;Gate.
  
  In Steins;Gate, Okabe Rintarou needs to complete n tasks (1 ≤ n ≤ 2 ⋅ 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, …, 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.
  
  Input
  
  The first line contains an integer n (1 ≤ n ≤ 2 ⋅ 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 ≤ a_k < b_k ≤ 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 ≤ t ≤ 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, …, s_t — (1 ≤ s_k ≤ n, the numbers s_1, s_2, …, s_t are distinct) — the set s of tasks.
  
  Output
  
  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.
  
  Examples
  
  Input
  
  
  2
  1 4
  2 3
  2
  1 2
  
  
  Output
  
  
  3
  
  
  Input
  
  
  2
  1 4
  2 3
  1
  1
  
  
  Output
  
  
  2
  
  
  Input
  
  
  1
  1 2
  1
  1
  
  
  Output
  
  
  1
  
  
  Input
  
  
  6
  10 12
  3 7
  4 6
  2 9
  5 8
  1 11
  3
  2 4 6
  
  
  Output
  
  
  17
  
  
  Input
  
  
  16
  31 32
  3 26
  17 19
  4 24
  1 28
  15 21
  12 16
  18 29
  20 23
  7 8
  11 14
  9 22
  6 30
  5 10
  25 27
  2 13
  6
  3 8 2 5 12 11
  
  
  Output
  
  
  138
  
  Note
  
  For the first sample, all tasks need to be completed in order for the funny scene to occur.
  
  Initially, the time is 0. Nothing happens until time 3, when Okabe realizes that he should have done the 2-nd task at time 2. He then time travels to time 2 and completes the task.
  
  As the task is done now, he does not time travel again when the time is again 3. However, at time 4, he travels to time 1 to complete the 1-st task.
  
  This undoes the 2-nd task. This means that the 2-nd task is not currently completed, meaning that the funny scene will not occur at this point even though the 1-st task is currently completed and Okabe had previously completed the 2-nd task.
  
  Once it is again time 3 he travels back to time 2 once more and does the 2-nd task again.
  
  Now all tasks are complete, with Okabe having time travelled 3 times.
  
  The second sample has the same tasks for Okabe to complete. However, this time the funny scene only needs the first task to be completed in order to occur. From reading the above sample you can see that this occurs once Okabe has time travelled 2 times.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int M = 1e9 + 7;
  pair<int, int> a[200004];
  int n, b[200004], m, bit[400004], dp[200004], ans, lst = 1;
  bool v[200004];
  void upd(int i, int x) {
    for (; i; i -= i & -i) bit[i] = (bit[i] + x) % M;
  }
  int got(int i) {
    int ans = 0;
    for (; i <= n + n; i += i & -i) ans = (ans + bit[i]) % M;
    return ans;
  }
  int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
      scanf("%d%d", &a[i].second, &a[i].first), b[i] = a[i].first;
    sort(a + 1, a + n + 1);
    scanf("%d", &m);
    for (int i = 1; i <= m; i++) {
      int x;
      scanf("%d", &x), x = b[x];
      v[lower_bound(a + 1, a + n + 1, make_pair(x, 0)) - a] = 1;
    }
    upd(n + n, 1);
    for (int i = 1; i <= n; i++) {
      dp[i] = got(a[i].second);
      upd(a[i].second, dp[i]);
    }
    for (int i = n; i; i--) {
      upd(a[i].second, M - dp[i]);
      if (v[i] && lst <= a[i].second) {
        ans = (ans + got(lst)) % M;
        lst = a[i].second;
      }
    }
    printf("%d", ans);
  }
  