Codeforces Round 735 (Div. 2) |
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Finished |

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A. Cherry

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputYou are given $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$. Find the maximum value of $$$max(a_l, a_{l + 1}, \ldots, a_r) \cdot min(a_l, a_{l + 1}, \ldots, a_r)$$$ over all pairs $$$(l, r)$$$ of integers for which $$$1 \le l < r \le n$$$.

Input

The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output

For each test case, print a single integer — the maximum possible value of the product from the statement.

Example

Input

4 3 2 4 3 4 3 2 3 1 2 69 69 6 719313 273225 402638 473783 804745 323328

Output

12 6 4761 381274500335

Note

Let $$$f(l, r) = max(a_l, a_{l + 1}, \ldots, a_r) \cdot min(a_l, a_{l + 1}, \ldots, a_r)$$$.

In the first test case,

- $$$f(1, 2) = max(a_1, a_2) \cdot min(a_1, a_2) = max(2, 4) \cdot min(2, 4) = 4 \cdot 2 = 8$$$.
- $$$f(1, 3) = max(a_1, a_2, a_3) \cdot min(a_1, a_2, a_3) = max(2, 4, 3) \cdot min(2, 4, 3) = 4 \cdot 2 = 8$$$.
- $$$f(2, 3) = max(a_2, a_3) \cdot min(a_2, a_3) = max(4, 3) \cdot min(4, 3) = 4 \cdot 3 = 12$$$.

So the maximum is $$$f(2, 3) = 12$$$.

In the second test case, the maximum is $$$f(1, 2) = f(1, 3) = f(2, 3) = 6$$$.

Codeforces (c) Copyright 2010-2023 Mike Mirzayanov

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