#include <bits/stdc++.h>
long long I[200003];
long long c[77];
int main() {
  long long n, k, x;
  scanf("%lld %lld %lld", &n, &k, &x);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &I[i]);
    long long a = I[i];
    int f = 0;
    while (a) {
      if (a & 1) c[f]++;
      f++;
      a >>= 1;
    }
  }
  long long max = 0;
  for (int i = 1; i <= n; i++) {
    long long a = I[i];
    int f = 0;
    while (a) {
      if (a & 1) c[f]--;
      f++;
      a >>= 1;
    }
    a = I[i];
    for (int j = 1; j <= k; j++) a *= x;
    long long ca = a;
    f = 0;
    while (a) {
      if (a & 1) c[f]++;
      f++;
      a >>= 1;
    }
    long long tmp = 0;
    for (long long j = 0; j <= 63; j++)
      if (c[j]) tmp += (1LL << j);
    if (max < tmp) max = tmp;
    a = ca;
    f = 0;
    while (a) {
      if (a & 1) c[f]--;
      f++;
      a >>= 1;
    }
    a = I[i];
    f = 0;
    while (a) {
      if (a & 1) c[f]++;
      f++;
      a >>= 1;
    }
  }
  printf("%lld\n", max);
}

### Prompt

Please create a solution in cpp to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
long long I[200003];
long long c[77];
int main() {
  long long n, k, x;
  scanf("%lld %lld %lld", &n, &k, &x);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &I[i]);
    long long a = I[i];
    int f = 0;
    while (a) {
      if (a & 1) c[f]++;
      f++;
      a >>= 1;
    }
  }
  long long max = 0;
  for (int i = 1; i <= n; i++) {
    long long a = I[i];
    int f = 0;
    while (a) {
      if (a & 1) c[f]--;
      f++;
      a >>= 1;
    }
    a = I[i];
    for (int j = 1; j <= k; j++) a *= x;
    long long ca = a;
    f = 0;
    while (a) {
      if (a & 1) c[f]++;
      f++;
      a >>= 1;
    }
    long long tmp = 0;
    for (long long j = 0; j <= 63; j++)
      if (c[j]) tmp += (1LL << j);
    if (max < tmp) max = tmp;
    a = ca;
    f = 0;
    while (a) {
      if (a & 1) c[f]--;
      f++;
      a >>= 1;
    }
    a = I[i];
    f = 0;
    while (a) {
      if (a & 1) c[f]++;
      f++;
      a >>= 1;
    }
  }
  printf("%lld\n", max);
}

```