CF1100F.Ivan and Burgers

Ivan loves burgers and spending money. There are $n$ burger joints on the street where Ivan lives. Ivan has $q$ friends, and the $i$ -th friend suggested to meet at the joint $l_i$ and walk to the joint $r_i$ $(l_i \leq r_i)$ . While strolling with the $i$ -th friend Ivan can visit all joints $x$ which satisfy $l_i \leq x \leq r_i$ .

For each joint Ivan knows the cost of the most expensive burger in it, it costs $c_i$ burles. Ivan wants to visit some subset of joints on his way, in each of them he will buy the most expensive burger and spend the most money. But there is a small issue: his card broke and instead of charging him for purchases, the amount of money on it changes as follows.

If Ivan had $d$ burles before the purchase and he spent $c$ burles at the joint, then after the purchase he would have $d \oplus c$ burles, where $\oplus$ denotes the bitwise XOR operation.

Currently Ivan has $2^{2^{100}} - 1$ burles and he wants to go out for a walk. Help him to determine the maximal amount of burles he can spend if he goes for a walk with the friend $i$ . The amount of burles he spends is defined as the difference between the initial amount on his account and the final account.

The first line contains one integer $n$ ( $1 \leq n \leq 500\,000$ ) — the number of burger shops.

The next line contains $n$ integers $c_1, c_2, \ldots, c_n$ ( $0 \leq c_i \leq 10^6$ ), where $c_i$ — the cost of the most expensive burger in the burger joint $i$ .

The third line contains one integer $q$ ( $1 \leq q \leq 500\,000$ ) — the number of Ivan's friends.

Each of the next $q$ lines contain two integers $l_i$ and $r_i$ ( $1 \leq l_i \leq r_i \leq n$ ) — pairs of numbers of burger shops between which Ivan will walk.

Output $q$ lines, $i$ -th of which containing the maximum amount of money Ivan can spend with the friend $i$ .

In the first test, in order to spend the maximum amount of money with the first and third friends, Ivan just needs to go into the first burger. With a second friend, Ivan just go to the third burger.

In the second test for a third friend (who is going to walk from the first to the third burger), there are only 8 options to spend money — $0$ , $12$ , $14$ , $23$ , $12 \oplus 14 = 2$ , $14 \oplus 23 = 25$ , $12 \oplus 23 = 27$ , $12 \oplus 14 \oplus 23 = 20$ . The maximum amount of money it turns out to spend, if you go to the first and third burger — $12 \oplus 23 = 27$ .