AtCoder Regular Contest 146
B
Plus and AND
AtCoder
⌀ 1335
原题
← AC →
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Problem Statement

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You are given a sequence of $N$ non-negative integers: $A=(A_1,A_2,\dots,A_N)$. You may perform the following operation at most $M$ times (possibly zero):

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  • choose an integer $i$ such that $1 \le i \le N$ and add $1$ to $A_i$.
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Then, you will choose $K$ of the elements of $A$.

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Find the maximum possible value of the bitwise $\mathrm{AND}$ of the elements you choose.

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What is bitwise ${\rm AND}$? The bitwise ${\rm AND}$ of non-negative integers $A$ and $B$, $A\ \mathrm{AND}\ B$, is defined as follows: When $A\ {\rm AND}\ B$ is written in base two, the digit in the $2^k$'s place ($k \geq 0$) is $1$ if both of the digits in that place of $A$ and $B$ are $1$, and $0$ otherwise. For example, $3\ {\rm AND}\ 5 = 1$. (In base two, $011\ {\rm AND}\ 101 = 001$.) Generally, the bitwise ${\rm AND}$ of $k$ non-negative integers $p_1, p_2, p_3, \dots, p_k$ is defined as $(\dots ((p_1\ \mathrm{AND}\ p_2)\ \mathrm{AND}\ p_3)\ \mathrm{AND}\ \dots\ \mathrm{AND}\ p_k)$. We can prove that this value does not depend on the order of $p_1, p_2, p_3, \dots, p_k$.

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Constraints

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  • $1 \le K \le N \le 2 \times 10^5$
  • $0 \le M < 2^{30}$
  • $0 \le A_i < 2^{30}$
  • All values in input are integers.
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Input

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Input is given from Standard Input in the following format:

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N M K
A_1 A_2 \dots A_N

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Output

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Print the answer.

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样例

Input #1
4 8 2
1 2 4 8
Output #1
10
If you do the following, the bitwise $\mathrm{AND}$ of the elements you choose will be $10$. - Perform the operation on $A_3$ six times. Now, $A_3 = 10$. - Perform the operation on $A_4$ twice. Now, $A_4 = 10$. - Choose $A_3$ and $A_4$, whose bitwise $\mathrm{AND}$ is $10$. There is no way to choose two elements whose bitwise $\mathrm{AND}$ is greater than $10$, so the answer is $10$.
Input #2
5 345 3
111 192 421 390 229
Output #2
461
题目信息
题号
arc146_b
时间限制
5 s
内存限制
1024 MB

题面与样例版权归 AtCoder 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。