Problem:
Score : 500 points

Problem Statement

Given are N integers A_1,\ldots,A_N.

We will choose exactly K of these elements. Find the maximum possible product of the chosen elements.

Then, print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

Constraints

1 \leq K \leq N \leq 2\times 10^5

|A_i| \leq 10^9

Input

Input is given from Standard Input in the following format:

N K
A_1 \ldots A_N

Output

Print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

Sample Input 1

4 2
1 2 -3 -4

Sample Output 1

12

The possible products of the two chosen elements are 2, -3, -4, -6, -8, and 12, so the maximum product is 12.

Sample Input 2

4 3
-1 -2 -3 -4

Sample Output 2

1000000001

The possible products of the three chosen elements are -24, -12, -8, and -6, so the maximum product is -6.

We print this value modulo (10^9+7), that is, 1000000001.

Sample Input 3

2 1
-1 1000000000

Sample Output 3

1000000000

The possible products of the one chosen element are -1 and 1000000000, so the maximum product is 1000000000.

Sample Input 4

10 10
1000000000 100000000 10000000 1000000 100000 10000 1000 100 10 1

Sample Output 4

999983200

Be sure to print the product modulo (10^9+7).

Solution in PyPy3 (7.3.0) (judge verdict: Runtime Error)

import sys
from operator import mul
from functools import reduce
from collections import deque

n, k = map(int, input().split())
arr = list(map(int, input().split()))


MOD = 10**9 + 7

# negを奇数個選ばないといけないシチュエーションかどうか
neg = [x < 0 for x in arr]

if all(neg) and k % 2 == 1:
    arr.sort(reverse=True)
    print(reduce(mul, arr[:k]) % MOD)
    sys.exit()

arr_p = [x for x in arr if x >= 0]
arr_n = [x for x in arr if x < 0]

arr_p.sort(reverse=True)
arr_n.sort()

arr_p = deque(arr_p)
arr_n = deque(arr_n)

ans = 1
cnt = 0
while len(arr_p) >= 2 and len(arr_n) >= 2 and k // 2 * 2 >= cnt:
    a = arr_p.popleft()
    b = arr_p.popleft()
    c = arr_n.popleft()
    d = arr_n.popleft()
    if a * b <= c * d:
        ans *= c * d
        ans %= MOD
        arr_n.appendleft(c)
        arr_n.appendleft(d)
    else:
        ans *= a * b
        ans %= MOD
        arr_p.appendleft(a)
        arr_p.appendleft(b)
    cnt += 2

if k != cnt:
    ans *= reduce(mul, arr_p[:k - cnt])

print(ans % MOD)