B"El Psy Kongroo. Omkar is watching Steins;Gate. In Steins;Gate, Okabe Rintarou needs to complete n tasks ( 1 <= q n <= q 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 <= q n <= q 2 cdot 10^5 ) -- the number of tasks that Okabe needs to complete. n lines follow. Th"...