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 maxProductSubarrayOfSizeK(A, n, k): 

	# sorting given input array 
	A.sort() 

	product = 1

	if (A[n - 1] == 0 and (k & 1)): 
		return 0

	if (A[n - 1] <= 0 and (k & 1)) : 
		for i in range(n - 1, n - k + 1, -1): 
			product *= A[i] 
		return product 

	# else 
	# i is current left pointer index 
	i = 0

	# j is current right pointer index 
	j = n - 1


	if (k & 1): 
		product *= A[j] 
		j-= 1
		k-=1

	k >>= 1

	for itr in range( k) : 

		# product from left pointers 
		left_product = A[i] * A[i + 1] 

		# product from right pointers 
		right_product = A[j] * A[j - 1] 

		if (left_product > right_product) : 
			product *= left_product 
			i += 2
		
		else : 
			product *= right_product 
			j -= 2

	# Finally return product 
	return product 

# Driver Code 
if __name__ == "__main__": 
	
	n,k = map(int,input().split())
	l = list(map(int,input().split()))
	ans=1
	if n==k:
	    for i in l:
	        ans = ans*i
	    print(ans%1000000007)
	    exit()
	ans = maxProductSubarrayOfSizeK(l, n, k)
	ans = ans%1000000007
	print(ans)