Codeforces Round #320 (Div. 2) [Bayan Thanks-Round] D "Or" Game 枚举+前缀后缀

                                                                        D. "Or" Game

                                                                            time limit per test

                                                                                  2 seconds

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 as large as possible, where denotes the bitwise OR.

Find the maximum possible value of 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.

Sample test(s)

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 .

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.

题解：肯定是全部乘在一个数上

     用前缀后缀枚举就好

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<set>
using namespace std;
#define maxn 201000
__int64 a = 1 , b , c , d[maxn*2] , L[maxn*2] , R[maxn*2], k , x , n;

int main()
{
    cin >> n >> k >> x;

    for(int i = 1 ; i <= n ; i++)
        cin >> d[i] , L[i] = L[i-1]|d[i];

    for(int i = 1 ; i <= k ; i++)
        a *= x;

    for(int i = n ; i >= 1 ; i--)
        R[i] = R[i+1]|d[i];

    for(int i = 1 ; i <= n ; i++)
        b = max(b , (L[i-1]|R[i+1]|((__int64)a*d[i])));

    cout << b << endl;

    return 0;
}