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.
ls = raw_input().split()
n = int(ls[0])
x = int(ls[2])
k = int(ls[1])
mul = int(1)
for i in range(0, k):
    mul *= x
lst = raw_input().split()
pref = [int(lst[i]) for i in range(0, len(lst))]
suf = [int(lst[i]) for i in range(0, len(lst))]
for i in range(1, len(lst)):
    pref[i] = pref[i - 1] | int(lst[i])
    suf[len(lst) - 1 - i] = suf[len(lst) - i] | int(lst[len(lst) - i - 1])
ans = int(0)
for i in range(1, len(lst) - 1):
    ans = max(ans, pref[i - 1] | (int(lst[i]) * mul) | suf[i + 1])
if (len(lst) - 2 >= 0):
    ans = max(ans, pref[len(lst) - 2] | (int(lst[len(lst) - 1]) * mul))
    ans = max(ans, (int(lst[0]) * mul) | suf[1])
if (len(lst) == 1):
    ans = mul * pref[0]
print ans