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

        B. "Or" Game

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given n numbers a₁, a₂, ..., a_n. You can perform at mostk operations. For each operation you can multiply one of the numbers byx. We want to make as large as possible, where denotes the bitwise OR.

Find the maximum possible value of after performing at mostk 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 21 1 1

Output

3

Input

4 2 31 2 4 8

Output

79

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;}