A retail team stores its customers and their purchase history in two tables. For every customer, they want to look past the latest purchase and see the one before it.

Write a single SQL query that, for every customer who has placed at least two orders, returns that customer's second most recent order.

An order is "more recent" than another order of the same customer when its order_date is later. If two orders of the same customer share the same order_date, the one with the larger order_id is treated as the more recent of the two. Under this rule every customer's orders have a strict newest-to-oldest ranking, so "second most recent" is always a single order.

Schema

CREATE TABLE Customers (
    customer_id   INTEGER PRIMARY KEY,
    customer_name TEXT NOT NULL
);

CREATE TABLE Orders (
    order_id    INTEGER PRIMARY KEY,
    customer_id INTEGER NOT NULL,
    order_date  TEXT NOT NULL,
    amount      INTEGER NOT NULL
);

order_date is stored as text in YYYY-MM-DD form, so ordinary string comparison orders dates correctly.
amount is a whole number of currency units (no decimal part).
Every row in Orders refers to a customer_id that exists in Customers.
A customer may have zero, one, or many orders.

Result Format

Return exactly five columns, in this order: customer_id, customer_name, order_id, order_date, amount.

Exactly one row per qualifying customer.
Customers with fewer than two orders (including customers with no orders at all) must not appear.
Order the result by customer_id ascending. Because at most one row is produced per customer, customer_id alone is already a total order — no further tie-break is needed. (inferred — the source did not state whether an extra sort key was required; the query is judgeable without one because customer_id is unique in the result.)
If no customer has two or more orders, the query returns no rows. That is a valid result.

Constraints

1 <= number of rows in Customers <= 120
0 <= number of rows in Orders <= 500
0 <= amount <= 1000000000
customer_id is unique in Customers; order_id is unique in Orders.
All order_date values fall in the range 2020-01-01 .. 2029-12-31.

(inferred — the source stated 1000 customers and 5000 orders. Each test case is shipped to the grader as a literal SQL setup script, and a 5000-row INSERT exceeds the grader's per-question payload budget, so the bounds are lowered to values the grader can actually carry. The largest shipped case sits exactly at 120 customers and 500 orders.)

Example 1

Data:

INSERT INTO Customers (customer_id, customer_name) VALUES
    (1, 'Alice'),
    (2, 'Bob');
INSERT INTO Orders (order_id, customer_id, order_date, amount) VALUES
    (101, 1, '2024-03-10', 500),
    (102, 1, '2024-02-15', 300),
    (103, 1, '2024-01-05', 120),
    (104, 2, '2024-03-01', 400),
    (105, 2, '2024-02-20', 100);

Result:

1|Alice|102|2024-02-15|300
2|Bob|105|2024-02-20|100

Explanation: Alice's orders newest first are 101 (2024-03-10), 102 (2024-02-15) and 103 (2024-01-05), so her second most recent is 102. Bob has 104 (2024-03-01) then 105 (2024-02-20), so his second most recent is 105.

Example 2

Data:

INSERT INTO Customers (customer_id, customer_name) VALUES
    (1, 'Ghost'),
    (2, 'Active');
INSERT INTO Orders (order_id, customer_id, order_date, amount) VALUES
    (1, 2, '2024-04-02', 75),
    (2, 2, '2024-04-09', 40);

Result:

2|Active|1|2024-04-02|75

Explanation: Ghost has placed no orders and is dropped. Active has two orders; the newer is order 2 (2024-04-09), so the second most recent is order 1 (2024-04-02).

Example 3

Data:

INSERT INTO Customers (customer_id, customer_name) VALUES
    (3, 'Tie');
INSERT INTO Orders (order_id, customer_id, order_date, amount) VALUES
    (9, 3, '2024-06-01', 10),
    (8, 3, '2024-06-01', 20);

Result:

3|Tie|8|2024-06-01|20

Explanation: Both orders fall on 2024-06-01. The larger order_id (9) counts as the more recent one, so the second most recent is order 8.

Notes

Submit one SELECT statement. Your result set is compared row by row against the expected result, in the order your query returns them, with columns joined by |.
Do not print a header row; only the data rows are compared.
amount is an integer column, so return it as it is stored. Do not apply any numeric formatting.