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)

N, K = map(int, input().split())
mod = 10**9 + 7
plus = []
minus = []
for i in map(int, input().split()):
    if i > 0:
        plus.append(i)
    elif i < 0:
        minus.append(i)
plen = len(plus)
mlen = len(minus)
zlen = N - plen - mlen
if K > plen + mlen:
    print(0)
    import sys
    sys.exit()
plus.sort(reverse=True)
minus.sort()
pp = [1]
mp = [1]
for p in plus:
    pp.append((pp[-1]*p)%mod)
for m in minus:
    mp.append((mp[-1]*(-m))%mod)
ppm = [1]
mpm = [1]
for p in plus[::-1]:
    ppm.append((ppm[-1]*p)%mod)
for m in minus[::-1]:
    mpm.append((mpm[-1]*(-m))%mod)
ans = -float("inf")
# kこマイナスから選ぶ
for k in range(K+1):
    if mlen < k:
        continue
    if plen < K-k:
        continue
    # 最大を目指す
    if k % 2 == 0:
        tmp = mp[k] * pp[K-k]
        ans = max(ans, tmp)
    # 最小を目指す
    else:
        tmp = -(mpm[k] * ppm[K-k])
        ans = max(ans, tmp)
if ans < 0 and zlen > 0:
    print(0)
    import sys
    sys.exit()
print(ans%mod)