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.
#include <bits/stdc++.h>
using namespace std;
long long temp;
int main() {
  long long n, k, x;
  long long a[200005];
  cin >> n >> k >> x;
  long long i;
  for (i = 0; i < n; ++i) cin >> a[i];
  long long preOr[200005], sufOr[200005];
  long long xPower[15];
  xPower[0] = 1;
  for (i = 1; i <= k; ++i) xPower[i] = xPower[i - 1] * x;
  preOr[0] = a[0];
  for (i = 1; i < n; ++i) preOr[i] = preOr[i - 1] | a[i];
  sufOr[n - 1] = a[n - 1];
  for (i = n - 2; i >= 0; --i) sufOr[i] = sufOr[i + 1] | a[i];
  long long ans;
  long long maxAns = 0;
  for (i = 0; i < n; ++i) {
    long long before, after;
    if (i > 0)
      before = preOr[i - 1];
    else
      before = 0;
    if (i < n - 1)
      after = sufOr[i + 1];
    else
      after = 0;
    long long cur = a[i] * xPower[k];
    ans = before | cur | after;
    if (maxAns < ans) maxAns = ans;
  }
  cout << maxAns;
}
