CF1637E Best Pair
Description
You are given an array $ a $ of length $ n $ . Let $ cnt_x $ be the number of elements from the array which are equal to $ x $ . Let's also define $ f(x, y) $ as $ (cnt_x + cnt_y) \cdot (x + y) $ .
Also you are given $ m $ bad pairs $ (x_i, y_i) $ . Note that if $ (x, y) $ is a bad pair, then $ (y, x) $ is also bad.
Your task is to find the maximum value of $ f(u, v) $ over all pairs $ (u, v) $ , such that $ u \neq v $ , that this pair is not bad, and also that $ u $ and $ v $ each occur in the array $ a $ . It is guaranteed that such a pair exists.
Input Format
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 two integers $ n $ and $ m $ ( $ 2 \le n \le 3 \cdot 10^5 $ , $ 0 \le m \le 3 \cdot 10^5 $ ) — the length of the array and the number of bad pairs.
The second line of each test case contains $ n $ integers $ a_1, a_2, \ldots, a_n $ ( $ 1 \le a_i \le 10^9 $ ) — elements of the array.
The $ i $ -th of the next $ m $ lines contains two integers $ x_i $ and $ y_i $ ( $ 1 \le x_i < y_i \le 10^9 $ ), which represent a bad pair. It is guaranteed that no bad pair occurs twice in the input. It is also guaranteed that $ cnt_{x_i} > 0 $ and $ cnt_{y_i} > 0 $ .
It is guaranteed that for each test case there is a pair of integers $ (u, v) $ , $ u \ne v $ , that is not bad, and such that both of these numbers occur in $ a $ .
It is guaranteed that the total sum of $ n $ and the total sum of $ m $ don't exceed $ 3 \cdot 10^5 $ .
Output Format
For each test case print a single integer — the answer to the problem.
Explanation/Hint
In the first test case $ 3 $ , $ 6 $ , $ 7 $ occur in the array.
- $ f(3, 6) = (cnt_3 + cnt_6) \cdot (3 + 6) = (3 + 2) \cdot (3 + 6) = 45 $ . But $ (3, 6) $ is bad so we ignore it.
- $ f(3, 7) = (cnt_3 + cnt_7) \cdot (3 + 7) = (3 + 1) \cdot (3 + 7) = 40 $ .
- $ f(6, 7) = (cnt_6 + cnt_7) \cdot (6 + 7) = (2 + 1) \cdot (6 + 7) = 39 $ .
The answer to the problem is $ \max(40, 39) = 40 $ .