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Given 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.

Example 1:

Input: n = 1
Output: 1
Explanation: Unique order (P1, D1), Delivery 1 always is after of Pickup 1.


Example 2:

Input: n = 2
Output: 6
Explanation: All possible orders:
(P1,P2,D1,D2), (P1,P2,D2,D1), (P1,D1,P2,D2), (P2,P1,D1,D2), (P2,P1,D2,D1) and (P2,D2,P1,D1).
This is an invalid order (P1,D2,P2,D1) because Pickup 2 is after of Delivery 2.


Example 3:

Input: n = 3
Output: 90


Constraints:

• 1 <= n <= 500

Solution: Combination

Let dp[i] denote the number of valid sequence of i nodes.

For i-1 nodes, the sequence length is 2(i-1).
For the i-th nodes,
If we put Pi at index = 0, then we can put Di at 1, 2, …, 2i – 2 => 2i-1 options.
If we put Pi at index = 1, then we can put Di at 2,3,…, 2i – 2 => 2i – 2 options.

If we put Pi at index = 2i-1, then we can put Di at 2i – 1=> 1 option.
There are total (2i – 1 + 1) / 2 * (2i – 1) = i * (2*i – 1) options

dp[i] = dp[i – 1] * i * (2*i – 1)

or

dp[i] = 2n! / 2^n

## C++

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