Codeforces Round #320 (Div. 1) [Bayan Thanks-Round] -- B. "Or" Game （容斥定理）-优快云博客

探讨了通过执行特定操作来最大化一组整数的异或运算值的方法。介绍了如何利用容斥原理正向与反向求解异或运算，最终实现最优解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

You are given n numbers a₁, a₂, ..., a_n. You can perform at most k operations. For each operation you can multiply one of the numbers by x. We want to make $\text{[math]}$ as large as possible, where $\text{[math]}$ denotes the bitwise OR.

Find the maximum possible value of $\text{[math]}$ 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, ..., a_n (0 ≤ a_i ≤ 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

Input

4 2 3
1 2 4 8

Output

Note

大体题意：
给你n 个元素的集合！
可以进行k 个操作，每个操作可以给一个数乘以x
问最后的集合进行异或运算最大值是多少！

思路：
利用容斥定理，先正着求一遍异或运算
在反着求一遍。
最后枚举每一个数乘以 k 个x 连乘
用这个数的前面异或这个数在异或后面的数即可！
ll tmp = bef[i-1] | (a[i]*x) | aft[i+1];

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 200000 + 10;
typedef long long ll;
ll bef[maxn],aft[maxn],a[maxn];
int main(){
    ll n,k,x;
    scanf("%I64d%I64d%I64d",&n,&k,&x);
    for (int i = 1; i <= n; ++i){
        scanf("%I64d",&a[i]);
    }
    ll xx= 1;
    for (int i = 0; i < k; ++i)xx*=x;
    x=xx;
    for (int i = 1; i <= n; ++i){
        bef[i] =  bef[i-1] | a[i];
    }
    for (int i = n; i >= 1; --i){
        aft[i] = aft[i+1] | a[i];
    }
    ll ans = 0;
    for (int i = 1; i <= n; ++i){
        ll tmp = bef[i-1] | (a[i]*x) | aft[i+1];
        ans = max(ans,tmp);
    }
    printf("%I64d\n",ans);
    return 0;
}