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 int a[2000005];
map<long long int, int> mp;
int bx[100];
int main() {
  unsigned long long int n, i, j, k, l, m, x, q = 1, num = 1, ans = 0;
  scanf("%lld%lld%lld", &n, &k, &x);
  for (i = 0; i < n; i++) scanf("%ld", &a[i]);
  for (i = 0; i < k; i++) {
    q *= x;
  }
  for (i = 0; i < n; i++) {
    unsigned long long int quotient = a[i];
    j = 0;
    int temp[65] = {0};
    while (quotient != 0) {
      temp[j++] = quotient % 2;
      quotient = quotient / 2;
    }
    int l;
    int k = 0;
    for (l = 0; l < j; l++) {
      if (temp[l]) bx[l]++;
    }
  }
  int ct = 0;
  for (i = 0; i < n; i++) {
    num = 0;
    unsigned long long int quotient = a[i];
    j = 0;
    int temp[65] = {0};
    while (quotient != 0) {
      temp[j++] = quotient % 2;
      quotient = quotient / 2;
    }
    int l;
    int k = 0;
    for (l = 0; l < j; l++) {
      if (temp[l]) bx[l]--;
      temp[l] = 0;
    }
    quotient = q * a[i];
    j = 0;
    while (quotient != 0) {
      temp[j++] = quotient % 2;
      quotient = quotient / 2;
    }
    for (l = 0; l < j; l++) {
      if (temp[l]) bx[l]++;
    }
    num = 0;
    for (l = 0; l < 65; l++) {
      if (bx[l]) num += powl(2, l);
    }
    for (l = 0; l < j; l++) {
      if (temp[l]) bx[l]--;
      temp[l] = 0;
    }
    j = 0;
    quotient = a[i];
    while (quotient != 0) {
      temp[j++] = quotient % 2;
      quotient = quotient / 2;
    }
    for (l = 0; l < j; l++) {
      if (temp[l]) bx[l]++;
      temp[l] = 0;
    }
    ans = max(num, ans);
  }
  printf("%lld\n", ans);
  return 0;
}
