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;
long long mod_v(long long num) {
  if (num >= 0)
    return (num % 1000000007);
  else
    return (num % 1000000007 + 1000000007) % 1000000007;
}
long long fast_pow(long long b, long long p, long long md) {
  long long r = 1 % md;
  while (p > 0) {
    if (p % 2 == 1) r = (r * b) % md;
    p = p / 2;
    b = (b * b) % md;
  }
  return r;
}
int countBits(int n) {
  int sum = 0;
  while (n > 0) {
    if (n & 1 == 1) sum++;
    n = n >> 1;
  }
  return sum;
}
long long gcd(long long a, long long b) {
  a = abs(a);
  b = abs(b);
  while (b) {
    a = a % b;
    swap(a, b);
  }
  return a;
}
vector<int> prime;
char sieve[100000];
void primeSieve(int n) {
  sieve[0] = sieve[1] = 1;
  prime.push_back(2);
  for (int i = 4; i <= n; i += 2) sieve[i] = 1;
  int sqrtn = sqrt(n);
  for (int i = 3; i <= sqrtn; i += 2) {
    if (sieve[i] == 0) {
      for (int j = i * i; j <= n; j += 2 * i) sieve[j] = 1;
    }
  }
  for (int i = 3; i <= n; i += 2)
    if (sieve[i] == 0) prime.push_back(i);
}
int n, k, x;
long long arr[200005] = {0};
long long lor[200005] = {0};
long long ror[200005] = {0};
int main() {
  scanf("%d %d %d", &n, &k, &x);
  for (int i = 0; i < n; i++) scanf("%I64d", &arr[i]);
  lor[0] = arr[0];
  for (int i = 1; i < n; i++) {
    lor[i] = lor[i - 1] | arr[i];
  }
  ror[n - 1] = arr[n - 1];
  for (int i = n - 2; i >= 0; i--) {
    ror[i] = ror[i + 1] | arr[i];
  }
  long long ans = 0, temp;
  for (int i = 0; i < n; i++) {
    temp = arr[i];
    for (int j = 0; j < k; j++) temp *= x;
    if (i > 0) temp = temp | lor[i - 1];
    if (i < n - 1) temp = temp | ror[i + 1];
    ans = max(ans, temp);
  }
  printf("%I64d\n", ans);
  return 0;
}
