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;
constexpr int kMaxBits{64};
int n, k, x;
int bitsCnt[kMaxBits];
vector<long long> nums;
int main() {
  cin >> n >> k >> x;
  for (int i = 0; i < n; ++i) {
    int temp;
    cin >> temp;
    nums.push_back(temp);
  }
  long long mul = 1;
  for (int i = 0; i < k; ++i) mul *= x;
  for (int i = 0; i < n; ++i)
    for (int b = 0; b < kMaxBits; ++b)
      if (nums[i] & (((long long)1) << b)) ++bitsCnt[b];
  long long res = 0;
  for (int i = 0; i < n; ++i) {
    for (int b = 0; b < kMaxBits; ++b)
      if (nums[i] & (((long long)1) << b)) --bitsCnt[b];
    long long possibleNumber = nums[i] * mul;
    long long currentRes = 0;
    for (int b = 0; b < kMaxBits; ++b)
      if ((possibleNumber & (((long long)1) << b)) || bitsCnt[b] > 0)
        currentRes += (((long long)1) << b);
    res = max(res, currentRes);
    for (int b = 0; b < kMaxBits; ++b)
      if (nums[i] & (((long long)1) << b)) ++bitsCnt[b];
  }
  cout << res << endl;
  return 0;
}
