E. Best Pair
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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

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

For each test case print a single integer — the answer to the problem.

Example
Input
3
6 1
6 3 6 7 3 3
3 6
2 0
3 4
7 4
1 2 2 3 1 5 1
1 5
3 5
1 3
2 5
Output
40
14
15
Note

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 $$$\max(40, 39) = 40$$$.