You are given n numbers a1, a2, ..., an. You can perform at most k operations. For each operation you can multiply one of the numbers by x. We want to make <image> as large as possible, where <image> denotes the bitwise OR. 

Find the maximum possible value of <image> after performing at most k operations optimally.

Input

The first line contains three integers n, k and x (1 ≤ n ≤ 200 000, 1 ≤ k ≤ 10, 2 ≤ x ≤ 8).

The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).

Output

Output the maximum value of a bitwise OR of sequence elements after performing operations.

Examples

Input

3 1 2
1 1 1


Output

3


Input

4 2 3
1 2 4 8


Output

79

Note

For the first sample, any possible choice of doing one operation will result the same three numbers 1, 1, 2 so the result is <image>. 

For the second sample if we multiply 8 by 3 two times we'll get 72. In this case the numbers will become 1, 2, 4, 72 so the OR value will be 79 and is the largest possible result.
inp=raw_input()
n=int(inp.split()[0])
k=int(inp.split()[1])
x=int(inp.split()[2])
s=raw_input()
arr=map(int,s.split())
v1=[]
v2=[]
v1.append(arr[0])
v2.append(0)
for i in range(1,n):
	v1.append(v1[i-1] | arr[i])
	v2.append(0)
v2[n-1]=arr[n-1]
for i in range(n-2,-1,-1):
	v2[i]=v2[i+1] | arr[i]
ans=0
for i in range (0,n):
	num=arr[i]
	for ii in range(0,k):
		num=num*x
	if i==0:
	 	if n>1:
			ans=num | v2[1]
		else : 
			ans=max(ans,num)
	elif i==n-1:
	 	if n>=2:
		 	ans=max(ans,v1[n-2] | num)
		else :
			ans=max(ans,num)
	else :
		ans=max(ans,(v1[i-1] | num ) | v2[i+1])
print ans

