Codeforces Round 735 (Div. 2)
A
Cherry
Codeforces
800
原题
←B →
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You 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 $1 \le l \lt r \le n$.

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Input

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

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

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

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

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Output

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

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Note

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

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In the first test case,

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  • $f(1, 2) = max(a_1, a_2) \cdot min(a_1, a_2) = max(2, 4) \cdot min(2, 4) = 4 \cdot 2 = 8$.
  • $f(1, 3) = max(a_1, a_2, a_3) \cdot min(a_1, a_2, a_3) = max(2, 4, 3) \cdot min(2, 4, 3) = 4 \cdot 2 = 8$.
  • $f(2, 3) = max(a_2, a_3) \cdot min(a_2, a_3) = max(4, 3) \cdot min(4, 3) = 4 \cdot 3 = 12$.
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So the maximum is $f(2, 3) = 12$.

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

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样例

Input #1
4
3
2 4 3
4
3 2 3 1
2
69 69
6
719313 273225 402638 473783 804745 323328
Output #1
12
6
4761
381274500335
InputCopy OutputCopy
题目信息
题号
1554A
时间限制
1 s
内存限制
256 MB
标签
贪心

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