CF1326E.Bombs

You are given a permutation, $p_1, p_2, \ldots, p_n$ .

Imagine that some positions of the permutation contain bombs, such that there exists at least one position without a bomb.

For some fixed configuration of bombs, consider the following process. Initially, there is an empty set, $A$ .

For each $i$ from $1$ to $n$ :

• Add $p_i$ to $A$ .
• If the $i$ -th position contains a bomb, remove the largest element in $A$ .

After the process is completed, $A$ will be non-empty. The cost of the configuration of bombs equals the largest element in $A$ .

You are given another permutation, $q_1, q_2, \ldots, q_n$ .

For each $1 \leq i \leq n$ , find the cost of a configuration of bombs such that there exists a bomb in positions $q_1, q_2, \ldots, q_{i-1}$ .

For example, for $i=1$ , you need to find the cost of a configuration without bombs, and for $i=n$ , you need to find the cost of a configuration with bombs in positions $q_1, q_2, \ldots, q_{n-1}$ .

The first line contains a single integer, $n$ ( $2 \leq n \leq 300\,000$ ).

The second line contains $n$ distinct integers $p_1, p_2, \ldots, p_n$ ( $1 \leq p_i \leq n)$ .

The third line contains $n$ distinct integers $q_1, q_2, \ldots, q_n$ ( $1 \leq q_i \leq n)$ .

Print $n$ space-separated integers, such that the $i$ -th of them equals the cost of a configuration of bombs in positions $q_1, q_2, \ldots, q_{i-1}$ .

In the first test:

• If there are no bombs, $A$ is equal to $\{1, 2, 3\}$ at the end of the process, so the cost of the configuration is $3$ .
• If there is one bomb in position $1$ , $A$ is equal to $\{1, 2\}$ at the end of the process, so the cost of the configuration is $2$ ;
• If there are two bombs in positions $1$ and $2$ , $A$ is equal to $\{1\}$ at the end of the process, so the cost of the configuration is $1$ .

In the second test:

Let's consider the process for $i = 4$ . There are three bombs on positions $q_1 = 5$ , $q_2 = 2$ , and $q_3 = 1$ .

At the beginning, $A = \{\}$ .

• Operation $1$ : Add $p_1 = 2$ to $A$ , so $A$ is equal to $\{2\}$ . There exists a bomb in position $1$ , so we should delete the largest element from $A$ . $A$ is equal to $\{\}$ .
• Operation $2$ : Add $p_2 = 3$ to $A$ , so $A$ is equal to $\{3\}$ . There exists a bomb in position $2$ , so we should delete the largest element from $A$ . $A$ is equal to $\{\}$ .
• Operation $3$ : Add $p_3 = 6$ to $A$ , so $A$ is equal to $\{6\}$ . There is no bomb in position $3$ , so we do nothing.
• Operation $4$ : Add $p_4 = 1$ to $A$ , so $A$ is equal to $\{1, 6\}$ . There is no bomb in position $4$ , so we do nothing.
• Operation $5$ : Add $p_5 = 5$ to $A$ , so $A$ is equal to $\{1, 5, 6\}$ . There exists a bomb in position $5$ , so we delete the largest element from $A$ . Now, $A$ is equal to $\{1, 5\}$ .
• Operation $6$ : Add $p_6 = 4$ to $A$ , so $A$ is equal to $\{1, 4, 5\}$ . There is no bomb in position $6$ , so we do nothing.

In the end, we have $A = \{1, 4, 5\}$ , so the cost of the configuration is equal to $5$ .