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: Accepted)

import sys
sys.setrecursionlimit(300000)
from heapq import heappush, heappop

def I(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def MI0(): return map(lambda s: int(s) - 1, sys.stdin.readline().split())
def LMI(): return list(map(int, sys.stdin.readline().split()))
def LMI0(): return list(map(lambda s: int(s) - 1, sys.stdin.readline().split()))
MOD = 10 ** 9 + 7
INF = float('inf')


N, K = MI()
A = LMI()

if all(a < 0 for a in A) and K % 2 == 1:
    A.sort(reverse=True)
    ans = 1
    for a in A[:K]:
        ans = ans * a % MOD
    print(ans)
    exit(0)

p = []
n = []
for a in A:
    if a < 0:
        heappush(n, a)
    else:
        heappush(p, -a)

ans = 1
cnt = 0
while cnt < K:
    if cnt == K - 1:
        if len(p) > 0:
            ans = ans * -heappop(p) % MOD
        else:
            ans = ans * max(n) % MOD
        cnt += 1
        break
    if len(p) < 1:
        ans = ans * heappop(n) * heappop(n) % MOD
        cnt += 2
    elif len(n) < 2:
        ans = ans * -heappop(p) % MOD
        cnt += 1
    else:
        p0, p1 = -p[0], -p[1] if len(p) > 1 else 1
        if p0 * p1 > n[0] * n[1]:
            ans = ans * -heappop(p) % MOD
            cnt += 1
        else:
            ans = ans * heappop(n) * heappop(n) % MOD
            cnt += 2
print(ans)