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 inf = 0x3f;
const int INF = 0x3f3f3f3f;
const int MAXN = 2e6 + 10;
long long n, k, x;
long long arr[MAXN];
struct SegTree {
  long long sum[MAXN << 2];
  void Init() { memset(sum, 0, sizeof(sum)); }
  void Build(int l, int r, int v) {
    if (l == r) {
      sum[v] = arr[l];
      return;
    }
    int mid = (l + r) >> 1;
    Build(l, mid, v << 1);
    Build(mid + 1, r, v << 1 | 1);
    sum[v] = sum[v << 1] | sum[v << 1 | 1];
  }
  long long Query(int l, int r, int v, int L, int R) {
    if (l >= L && r <= R) {
      return sum[v];
    }
    long long ans = 0;
    int mid = (l + r) >> 1;
    if (L <= mid) ans |= Query(l, mid, v << 1, L, R);
    if (R > mid) ans |= Query(mid + 1, r, v << 1 | 1, L, R);
    return ans;
  }
} seg;
int main() {
  while (cin >> n >> k >> x) {
    for (int i = 0; i < n; i++) cin >> arr[i];
    seg.Init();
    seg.Build(0, n - 1, 1);
    long long t = 1, mx = 0;
    for (int i = 0; i < k; i++) t *= x;
    for (int i = 0; i < n; i++) {
      long long sum = 0;
      if (i - 1 >= 0) sum |= (long long)seg.Query(0, n - 1, 1, 0, i - 1);
      if (i + 1 < n) sum |= (long long)seg.Query(0, n - 1, 1, i + 1, n - 1);
      mx = max(mx, sum | (arr[i] * t));
    }
    cout << mx << endl;
  }
  return 0;
}
