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;
inline int readInt() {
  int sign = 1, x = 0, c = getc(stdin);
  while (c <= 32) {
    c = getc(stdin);
  }
  if (c == '-') {
    sign = -1;
    c = getc(stdin);
  }
  while ('0' <= c && c <= '9' && c != EOF) {
    x = (x << 1) + (x << 3) + c - '0';
    c = getc(stdin);
  }
  return x * sign;
}
inline void writeLong(long long x) {
  if (x < 0) {
    putc('-', stdout);
    x = -x;
  }
  char s[40];
  int n = 0;
  while (x || !n) {
    s[n++] = '0' + x % 10;
    x /= 10;
  }
  while (n--) {
    putc(s[n], stdout);
  }
}
bool des(int i, int j) { return i > j; }
bool asc(int i, int j) { return i < j; }
const int M = 200003;
int a[M], suf[M], pre[M];
int main(int argc, char const *argv[]) {
  int n = readInt(), k = readInt(), x = readInt();
  for (int i = 1; i <= n; ++i) a[i] = readInt();
  unsigned long long pow = 1;
  while (k--) pow *= x;
  unsigned long long ans = 0, res = 0;
  for (int i = 1; i <= n; ++i) pre[i] = a[i - 1] | pre[i - 1];
  for (int j = n; j > 0; --j) suf[j] = a[j + 1] | suf[j + 1];
  for (int i = 1; i <= n; ++i) ans = max(ans, pre[i] | a[i] * pow | suf[i]);
  writeLong(ans);
  putchar('\n');
  return 0;
}
