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
read = sys.stdin.read
large_p = 10**9 + 7
from collections import deque
from copy import deepcopy
from operator import mul
from functools import reduce

def main():
    n, k, *a = map(int, read().split())
    if n == 1:
        print(a[0] % large_p)
        sys.exit()
    elif n == 2:
        r = max(a[0], a[1], a[0] * a[1])
        print(r % large_p)
        sys.exit()
    a2 = [i for i in a if i > 0] #plus
    a3 = [-i for i in a if i < 0] #minus
    a2.sort()
    a3.sort(reverse=True)
    #res = deepcopy(a2[(n - k):])
    if k < len(a2):
        res = deepcopy(a2[:-k])
    else:
        res = deepcopy(a2)
    resd = deque(res)
    a3d = deque(a3)
    a2dd = deque()
    a3dd = deque()
    cnt = 0
    while cnt < k:
        if a3d and resd[-1] < a3d[0]:
            a2dd.append(resd.pop())
            t = a3d.popleft()
            resd.appendleft(t)
            a3dd.append(t)
            cnt += 1
        else:
            break
    r = reduce(mul, resd)
    if r < 0:
        r = r // a3dd[-1] * -1
        r *= a2dd[-1]
    r = r % large_p
    if r < 0:
        r += large_p
    print(r)


if __name__ == '__main__':
    main()