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;
int N, k;
int pot;
long long a[200005], x;
int bitcnt[34];
int main() {
  int aux;
  scanf("%d%d", &N, &k);
  scanf("%d", &aux);
  x = aux;
  for (int i = 0; i < N; i++) {
    scanf("%I64d", &a[i]);
  }
  for (int i = 1; i < k; i++) x *= aux;
  for (int i = 0; i < N; i++) {
    long long tmp = 1;
    for (int bit = 0; bit < 33; bit++) {
      if (a[i] & tmp) bitcnt[bit]++;
      tmp <<= 1;
    }
  }
  long long ans = 0;
  for (int i = 0; i < N; i++) {
    long long tmp = 1;
    for (int bit = 0; bit < 33; bit++) {
      if (a[i] & tmp) bitcnt[bit]--;
      tmp <<= 1;
    }
    long long tmpans = a[i] * x;
    tmp = 1;
    for (int bit = 0; bit < 33; bit++) {
      if (bitcnt[bit]) tmpans |= tmp;
      tmp <<= 1;
    }
    ans = max(ans, tmpans);
    tmp = 1;
    for (int bit = 0; bit < 33; bit++) {
      if (a[i] & tmp) bitcnt[bit]++;
      tmp <<= 1;
    }
  }
  printf("%I64d\n", ans);
}
