Problem Statement

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):

• choose an integer $i$ such that $1 \\le i \\le N$ and add $1$ to $A_i$.

Then, you will choose $K$ of the elements of $A$.

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

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

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

Input

Input is given from Standard Input in the following format:

$N$ $M$ $K$ $A_1$ $A_2$ $\\dots$ $A_N$

Output

Print the answer.

Sample Input 1

4 8 2
1 2 4 8

Sample 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 $10$.

Sample Input 2

5 345 3
111 192 421 390 229

Sample Output 2

461