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 heapq as hp
import bisect as bs
import sys

def cal(array):
  ans = 1
  for i in range(len(array)):
    ans *= array[i]
    ans %= 10**9+7
    
  return ans

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

a = sorted(list(map(int, input().split())))
index = bs.bisect_left(a, 0)

c = a[:index]
index = bs.bisect_right(a, 0)

b = a[index:]

if len(b) + len(c) < k:
  print(0)
  sys.exit()
  
if len(a) == k:
  print(cal(k))
  sys.exit()
    

if len(b) == 0 and k % 2 != 0:
  ans = 1
  c = list(reversed(c))
  
  print(cal(c[:k]))
  sys.exit()
  
  
if len(b) < k:
  if (k - len(b)) % 2 == 0:
    for i in range(k-len(b)):
      b.append(hp.heappop(c) * (-1))
      
    print(cal(b))
    sys.exit()
  
  else:
    d = []
    c = reversed(c)
    for i in range(len(b)-1):
      d.append(c[i])
      
    if (k-len(b)+1) > len(b):
      print(0)
      sys.exit()
    
    else:
      for i in range(k-len(b)+1):
        d.append(c[i])
        
      ans = 1
      for i in range(len(d)):
        ans *= d[i]
        ans %= 10 ** 9 + 7
      
      print(ans)
  
while True:
  if len(c) < 2:
    break
  
  b1 = hp.heappop(b)
  b2 = hp.heappop(b)
    
  c1 = hp.heappop(c)
  c2 = hp.heappop(c)
    
  
  if b1*b2 <= c1*c2:
    hp.heappush(b, c1)
    hp.heappush(b, c2)
  else:
    break
    
ans = 1
for i in range(len(b)):
  ans *= b[i]
  ans %= 10 ** 9 + 7
  
print(ans)