Codeforces 578B Or Game (prefix + and greedy)

Codeforces Round #320 (Div. 1) [Bayan Thanks-Round]

You are given \(n\) numbers \(a_1, a_2, ..., a_n\). You can perform at most \(k\) operations. For each operation you can multiply one of the numbers by \(x\). We want to make \(a_1 | a_2 | ... | a_n\) as large as possible, where \(|\) denotes the bitwise OR.

Find the maximum possible value of \(a_1 | a_2 | ... | a_n\) 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 \(a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9)\).

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 \(1 | 1 | 2 = 3\).

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.

Solution

The meaning of problems

Given \ (n-\) number \ (A_1 ... A_N \) , may be \ (K \) operations, each time a number can be multiplied by any given \ (X \) , seeking \ (a_1 | a_2 | ... | a_n \) up to much.

answer

Prefix and greed

Since \ (the X-\ GE 2 \) , and therefore a number multiplied by \ (x \) after an increase in the number of bits inevitable.

Since the operation is or \ (0 | 1 \) if the answer will increase, so let \ (k \) a \ (x \) multiplied by the number on the same line.

Prefixes and suffixes and calculate and then multiplied by the number of violent enumerate each \ (the X-^ k \) , you can find the maximum value.

Code

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
typedef long long ll;

ll a[maxn], s1[maxn], s2[maxn];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n, k, x;
    cin >> n >> k >> x;
    for(int i = 1; i <= n; ++i) {
        cin >> a[i];
        s1[i] = s1[i - 1] | a[i];
    }
    for(int i = n; i >= 1; --i) {
        s2[i] = s2[i + 1] | a[i];
    }
    ll ans = 0;
    ll tmp = x;
    for(int i = 2; i <= k; ++i) {
        tmp *= x;
    }
    for(int i = 1; i <= n; ++i) {
        ans = max(ans, s1[i - 1] | tmp * a[i] | s2[i + 1]);
    }
    cout << ans << endl;
    return 0;
}