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;
const int maxn = 200010;
int n, k, x;
int a[maxn];
long long b[maxn];
int c[70];
long long quipow(int x, int k) {
  long long ret = 1;
  while (k) {
    if (k & 1) ret = ret * x;
    x = x * x;
    k >>= 1;
  }
  return ret;
}
int main() {
  cin >> n >> k >> x;
  long long p = 1;
  for (int i = (1); i <= (n); i++) {
    scanf("%d", &a[i]);
    while (a[i] >= p) p <<= 1;
  }
  p >>= 1;
  if (p == 0)
    puts("0");
  else {
    long long ans = 0;
    int cnt = 0;
    memset((c), (0), sizeof(c));
    for (int i = (1); i <= (n); i++) {
      ans |= a[i];
      b[i] = a[i] * quipow(x, k);
      for (int j = 0, t = a[i]; t; j++) {
        if (t & 1) c[j]++;
        t >>= 1;
      }
    }
    long long maxa = 0, tmp;
    for (int i = (1); i <= (n); i++) {
      for (int j = 0, t = a[i]; t; j++) {
        if (t & 1) c[j]--;
        t >>= 1;
      }
      tmp = b[i];
      for (int j = (0); j <= (64); j++) {
        if (c[j]) tmp |= (1ll << j);
      }
      maxa = ((maxa) > (tmp) ? (maxa) : (tmp));
      for (int j = 0, t = a[i]; t; j++) {
        if (t & 1) c[j]++;
        t >>= 1;
      }
    }
    cout << maxa << endl;
  }
  return 0;
}
