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 $t$ ($1 \le t \le 10000$) — 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.


-----Examples-----

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

Source: codeforces | Difficulty: HARD
Tags: Sorting

Implementation

Complete search


Solutions:
import sys

def solve():
	inp = sys.stdin.readline
	(n, m) = map(int, inp().split())
	a = list(map(int, inp().split()))
	bad = set()
	for i in range(m):
		bad.add(tuple(sorted(map(int, inp().split()))))
	a.sort(reverse=True)
	cnt = []
	i = 0
	while i < n:
		j = i + 1
		while j < n:
			if a[j] != a[i]:
				break
			j += 1
		c = j - i
		while c >= len(cnt):
			cnt.append([])
		cnt[c].append(a[i])
		i = j
	best = -1
	for i in range(len(cnt)):
		for u in cnt[i]:
			for j in range(i + 1):
				for v in cnt[j]:
					if u < v:
						if (u, v) not in bad:
							best = max(best, (u + v) * (i + j))
							break
					elif u > v:
						if (v, u) not in bad:
							best = max(best, (u + v) * (i + j))
							break
	print(best)

def main():
	for i in range(int(sys.stdin.readline())):
		solve()
main()


import math, bisect, heapq, random, sys, itertools
input = sys.stdin.readline
ints = lambda : list(map(int, input().split()))
alpha = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
p = 10 ** 9 + 7
ANS = []
for t in range(int(input())):
	(n, m) = ints()
	a = ints()
	bad = set()
	for i in range(m):
		(x, y) = ints()
		bad.add((x, y))
		bad.add((y, x))
	store = {}
	for i in a:
		if i not in store:
			store[i] = 0
		store[i] += 1
	count = {}
	for i in store:
		if store[i] not in count:
			count[store[i]] = []
		count[store[i]].append(i)
	for i in count:
		count[i].sort(reverse=True)
	cntx = sorted(count.keys())
	ans = 0
	for i in range(len(cntx)):
		ctx = cntx[i]
		for x in count[ctx]:
			for j in range(i + 1):
				cty = cntx[j]
				for y in count[cty]:
					if x != y and (x, y) not in bad:
						ans = max(ans, (x + y) * (ctx + cty))
						break
	ANS.append(ans)
print(*ANS, sep='\n')