n orders, each order consist in pickup and delivery services.
Count all valid pickup/delivery possible sequences such that delivery(i) is always after of pickup(i).
Since the answer may be too large, return it modulo 10^9 + 7.
Input: n = 1
We assumed it already have
n - 1 pair, and we have to place the final pair. The number of we can place is
2 * (n - 1) + 1 = 2*n - 1, so for the last, it should be
2 * n, which add one more section we can use due to the operation.
// time:O(n), space:O(1)