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
class Solution:
    def countOrders(self, n: int) -> int:
        
        
        
        dp = [1 for i in range(n)]
        
        for i in range(1,n):
            sequence = i+1
            length = 2*(i)
            temp = sum([j for j in range(1,length+2)])
            dp[i] = temp*dp[i-1]
        
        print(dp)
        return dp[-1]%(10**9+7)
class Solution:
    def countOrders(self, n: int) -> int:
        
        
        
        dp = [1 for i in range(n)]
        
        for i in range(1,n):
            sequence = i+1
            length = 2*(i)
            temp = sum([j for j in range(1,length+2)])
            dp[i] = temp*dp[i-1]
        
        print(dp)
        return dp[-1]%(10**9+7)