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 bool isLower(char ch) { return (ch >= 'a' && ch <= 'z'); }
inline bool isUpper(char ch) { return (ch >= 'A' && ch <= 'Z'); }
inline bool isDigit(char ch) { return (ch >= '0' && ch <= '9'); }
inline bool isVowel(char ch) {
  ch = tolower(ch);
  return (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u');
}
void printSet(int vS) {
  printf("S = %2d = ", vS);
  stack<int> st;
  while (vS) st.push(vS % 2), vS /= 2;
  while (!st.empty()) printf("%d", st.top()), st.pop();
  printf("\n");
}
long long int ar[2000006];
int main() {
  int n, k, x;
  cin >> n >> k >> x;
  long long int mul = x;
  for (int i = 2; i <= k; i++) {
    mul *= x;
  }
  int mxInd = 0, mxValue = -111;
  ;
  for (int i = 1; i <= n; i++) cin >> ar[i];
  long long int pref[n + 5];
  long long int suf[n + 5];
  pref[0] = 0;
  for (int i = 1; i <= n; i++) pref[i] = pref[i - 1] | ar[i];
  suf[n + 1] = 0;
  for (int i = n; i > 0; i--) suf[i] = suf[i + 1] | ar[i];
  long long int res = 0;
  for (int i = 1; i <= n; i++) {
    res = max(res, (ar[i] * mul) | pref[i - 1] | suf[i + 1]);
  }
  cout << res << endl;
  return 0;
}
