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 MAXPOW = 20;
const int MAXA = 1048577;
const int MAXN = 100009;
const int MAXM = 100009;
const int BITS = 32;
vector<int> getBitmask(int number) {
  vector<int> bitmask;
  for (int i = 0; i < BITS; ++i) {
    bitmask.push_back(number & 1);
    number >>= 1;
  }
  return bitmask;
}
int toNumber(const vector<int>& bitmask) {
  int res = 0;
  for (int i = bitmask.size(); i >= 0; --i) {
    res *= 2;
    res += bitmask[i] > 0;
  }
  return res;
}
vector<int> subBitmask(vector<int> sumBitmask, const vector<int>& bitmask) {
  for (int i = 0; i < BITS; ++i) {
    sumBitmask[i] -= bitmask[i];
  }
  return sumBitmask;
}
int main() {
  ios::sync_with_stdio(false);
  int n, k, x;
  cin >> n >> k >> x;
  vector<int> a(n);
  vector<vector<int>> bitmasks;
  vector<int> sumBitmasks(BITS, 0);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    auto bitmask = getBitmask(a[i]);
    bitmasks.push_back(bitmask);
    for (int i = 0; i < BITS; ++i) {
      sumBitmasks[i] += bitmask[i];
    }
  }
  long long xPow = 1;
  for (int i = 0; i < k; ++i) {
    xPow *= x;
  }
  long long maxResult = 0;
  for (int i = 0; i < n; ++i) {
    vector<int> withoutBitmask = subBitmask(sumBitmasks, bitmasks[i]);
    long long number = toNumber(bitmasks[i]);
    long long rest = toNumber(withoutBitmask);
    number *= xPow;
    long long result = number | rest;
    if (result > maxResult) {
      maxResult = result;
    }
  }
  cout << maxResult << endl;
  return 0;
}
