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G - Partial Xor Enumeration

Score : $600$ points

Problem Statement

For a sequence of non-negative integers $(a _ 1,a _ 2,\ldots,a _ n)$, let us define its $\operatorname{xor}$ as the integer $X$ such that, for all non-negative integer $j$:

• the $2^j$s place of $X$ is $1$ if and only if there is an odd number of elements among $a _ 1,\ldots,a _ n$ whose $2^j$s place is $1$.

You are given a sequence of non-negative integers $A=(A _ 1,A _ 2,\ldots,A _ N)$ of length $N$.

Let $\lbrace s _ 1,s _ 2,\ldots,s _ k\rbrace\ (s _ 1\lt s _ 2\lt\cdots\lt s _ k)$ be the set of all non-negative integers that can be the $\operatorname{xor}$ of a not-necessarily-contiguous (possibly empty) subsequence of $A$.

Given integers $L$ and $R$, print $s _ L,s _ {L+1},\ldots,s _ R$ in this order.

Constraints

• $1\leq N\leq2\times10^5$
• $0\leq A _ i\lt2^{60}\ (1\leq i\leq N)$
• $1\leq L\leq R\leq k$
• $R-L\leq2\times10^5$
• All values in the input are integers.

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Input

The input is given from Standard Input in the following format:

$N$ $L$ $R$

$A _ 1$ $A _ 2$ $\ldots$ $A _ N$

Output

Print $s _ i\ (L\leq i\leq R)$ in ascending order of $i$, separated by spaces.

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4 1 8
2 21 17 21

0 2 4 6 17 19 21 23

There are $14$ not-necessarily-contiguous subsequences of $A$: $(),(2),(17),(21),(2,17),(2,21),(17,21),(21,17),(21,21),(2,17,21),(2,21,17),(2,21,21),(21,17,21)$, and $(2,21,17,21)$, whose $\operatorname{xor}$s are as follows.

• The $\operatorname{xor}$ of $()$ is $0$.
• The $\operatorname{xor}$ of $(2)$ is $2$.
• The $\operatorname{xor}$ of $(17)$ is $17$.
• The $\operatorname{xor}$ of $(21)$ is $21$.
• The $\operatorname{xor}$ of $(2,17)$ is $19$.
• The $\operatorname{xor}$ of $(2,21)$ is $23$.
• The $\operatorname{xor}$ of $(17,21)$ is $4$.
• The $\operatorname{xor}$ of $(21,17)$ is $4$.
• The $\operatorname{xor}$ of $(21,21)$ is $0$.
• The $\operatorname{xor}$ of $(2,17,21)$ is $6$.
• The $\operatorname{xor}$ of $(2,21,17)$ is $6$.
• The $\operatorname{xor}$ of $(2,21,21)$ is $2$.
• The $\operatorname{xor}$ of $(21,17,21)$ is $17$.
• The $\operatorname{xor}$ of $(2,21,17,21)$ is $19$.

Therefore, the set of all non-negative integers that can be the $\operatorname{xor}$ of a subsequence of $A$ is $\lbrace0,2,4,6,17,19,21,23\rbrace$.

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4 3 7
2 21 17 21

4 6 17 19 21

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5 1 1
0 0 0 0 0

0

$A$ may contain the same value.

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6 21 21
88 44 82 110 121 80

41

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12 26 48
19629557415 14220078328 11340722069 30701452525 22333517481 720413777 11883028647 20926361028 24376768297 720413777 27999065315 13558621130

13558621130 14220078328 14586054825 15518998043 15970974282 16379590008 17091531049 17412316967 17836964726 18263536708 18965057557 19629557415 20282860278 20926361028 21302757781 21908867832 22333517481 22893781403 23595304394 23723463544 24376768297 24885524507 25261923402

Note that the input and output may not fit into a $32$-$\operatorname{bit}$ integer type.