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: Wrong Answer)

def prod(array):
    res = 1
    for a, _ in array:
        res *= a
        res %= MOD
    return res

MOD = 10**9 + 7
N, K = map(int, input().split())
if N==K:
    res = 1
    for a in map(int, input().split()):
        res *= a
        res %= MOD
    print(res)
    exit()
A = []
for a in map(int, input().split()):
    A.append((abs(a), a < 0))
A.sort(reverse=True)
sgn = 0
s0 = -1
lastA = -1
for i, p in enumerate(A[:K]):
    a, s = p
    sgn += s
    if s != s0:
        lastA = i-1
    s0 = s
lasts = s0
if sgn + sum(s for _, s in A[K:])==N:
    if K%2==1:
        print(-prod(A[-K:])%MOD)
        exit()
    else:
        print(prod(A[:K])%MOD)
        exit()
if sgn%2==0:
    print(prod(A[:K])%MOD)
    exit()
if lasts != A[K][1]:
    res = prod(A[:K-1])*A[K][0]
else:
    firstA = None
    for i, p in enumerate(A[K:],start=K):
        a, s = p
        if s != lasts:
            firstA = i
            break
    res1 = prod(A[:K-1])*A[firstA][0] if firstA is not None else -float("inf")
    res2 = prod(A[:lastA])*prod(A[lastA+1:K+1]) if lastA != -1 else -float("inf")
    res = max(res1, res2)
print(res%MOD)
"""
6 4
5 -3 -3 -3 -3 -2

5 3
-1 -2 -3 -4 -5
"""