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;
bool dbg = 0;
long long N, K, P;
vector<vector<long long> > inp;
long long dp[200005][15];
long long Solve(int pos, int Left) {
  if (pos == N) {
    return (long long)0;
  } else {
    if (dp[pos][Left] != (long long)-1) {
      return dp[pos][Left];
    }
    long long Ans = 0;
    for (int n = 0; n <= Left; ++n) {
      Ans = max(Ans, inp[pos][n] | Solve(pos + 1, Left - n));
    }
    return (dp[pos][Left] = Ans);
  }
}
long long pref[200005];
long long suff[200005];
int main() {
  cin >> N >> K >> P;
  inp.resize(N + 5);
  long long x;
  memset(dp, (long long)-1, sizeof(dp));
  ;
  vector<long long> a;
  a.push_back((long long)-1);
  for (int n = 1; n <= N; ++n) {
    scanf("%lld", &x);
    a.push_back(x);
  }
  long long mul = 1;
  for (int n = 0; n < K; ++n) {
    mul *= P;
  }
  for (int n = 1; n <= N; ++n) {
    pref[n] = pref[n - 1] | a[n];
  }
  for (int n = N; n >= 1; --n) {
    suff[n] = suff[n + 1] | a[n];
  }
  long long Ans = 0;
  for (int n = 1; n <= N; ++n) {
    Ans = max(Ans, pref[n - 1] | (a[n] * mul) | suff[n + 1]);
  }
  cout << Ans << endl;
  return 0;
}
