CF1866F.Freak Joker Process

After the success of the basketball teams formed and trained by Pak Chanek last year (Basketball Together), Pak Chanek wants to measure the performance of each player that is considered as a superstar.

There are $N$ superstar players that have been trained by Pak Chanek. At the end of the season, some calculations will be made on the performance of the $N$ players using an international method. Each player has two values $A_i$ and $B_i$ where each represents the offensive and defensive value of that player.

Define $\text{RankA}(i)$ as the offensive ranking of the $i$-th player, whose value is $c+1$ with $c$ here representing the number of $j$ ($1 \leq j \leq N$) such that $A_j \gt A_i$. Define $\text{RankB}(i)$ as the defensive ranking of the $i$-th player, whose value is $c+1$ with $c$ here representing the number of $j$ ($1 \leq j \leq N$) such that $B_j \gt B_i$.

Define $\text{RankOverall}(i)$ as the overall ranking of the $i$-th player, whose value is $c+1$ with $c$ here representing the number of $j$ ($1 \leq j \leq N$) such that $\text{RankA}(j)+\text{RankB}(j) \lt \text{RankA}(i)+\text{RankB}(i)$.

During the next $Q$ days, exactly one event will happen on each day. Each event is one of the three following possibilities:

• 1 k c – If $c$ is +, then $A_k$ increases by $1$. If $c$ is -, then $A_k$ decreases by $1$. ($1\leq k\leq N$; $c$ is + or -)
• 2 k c – If $c$ is +, then $B_k$ increases by $1$. If $c$ is -, then $B_k$ decreases by $1$. ($1\leq k\leq N$; $c$ is + or -)
• 3 k – Pak Chanek wants to know the value of $\text{RankOverall}(k)$ at that moment. ($1\leq k\leq N$)

在 Pak Chanek 去年成功组建并训练篮球队（“同心篮球”）之后，Pak Chanek 希望评估每位被视作超级明星的球员的表现。

共有 $N$ 名由 Pak Chanek 训练的超级明星球员。赛季末，将采用一种国际通用方法，对这 $N$ 名球员的表现进行若干计算。每名球员有两个数值 $A_i$ 和 $B_i$，分别代表该球员的进攻能力值和防守能力值。

定义 $\text{RankA}(i)$ 为第 $i$ 名球员的进攻排名，其值为 $c+1$，其中 $c$ 表示满足 $A_j \gt A_i$ 的下标 $j$ 的个数（$1 \leq j \leq N$）。
定义 $\text{RankB}(i)$ 为第 $i$ 名球员的防守排名，其值为 $c+1$，其中 $c$ 表示满足 $B_j \gt B_i$ 的下标 $j$ 的个数（$1 \leq j \leq N$）。

定义 $\text{RankOverall}(i)$ 为第 $i$ 名球员的综合排名，其值为 $c+1$，其中 $c$ 表示满足 $\text{RankA}(j)+\text{RankB}(j) \lt \text{RankA}(i)+\text{RankB}(i)$ 的下标 $j$ 的个数（$1 \leq j \leq N$）。

接下来的 $Q$ 天中，每天恰好发生一个事件。每个事件为以下三种类型之一：

• 1 k c — 若 $c$ 为 +，则 $A_k$ 增加 $1$；若 $c$ 为 -，则 $A_k$ 减少 $1$。（$1\leq k\leq N$；$c$ 为 + 或 -）
• 2 k c — 若 $c$ 为 +，则 $B_k$ 增加 $1$；若 $c$ 为 -，则 $B_k$ 减少 $1$。（$1\leq k\leq N$；$c$ 为 + 或 -）
• 3 k — Pak Chanek 想知道此时 $\text{RankOverall}(k)$ 的值。（$1\leq k\leq N$）

The first line contains a single integer $N$ ($1\leq N\leq10^5$) — the number of superstar players.

The second line contains $N$ integers $A_1, A_2, A_3 \ldots, A_N$ ($1 \leq A_i \leq 10^5$) — the offensive value of each player.

The third line contains $N$ integers $B_1, B_2, B_3 \ldots, B_N$ ($1 \leq B_i \leq 10^5$) — the defensive value of each player.

The fourth line contains a single integer $Q$ ($1\leq Q\leq10^5$) — the number of events.

The $j$-th of the next $Q$ lines contains the $j$-th event as described. At any moment, each value of $A_i$ and $B_i$ is always between $1$ and $10^5$ inclusive. There is at least one event of type $3$.

第一行包含一个整数 $N$（$1\leq N\leq10^5$）——超级明星球员的数量。

第二行包含 $N$ 个整数 $A_1, A_2, A_3 \ldots, A_N$（$1 \leq A_i \leq 10^5$）——每位球员的进攻值。

第三行包含 $N$ 个整数 $B_1, B_2, B_3 \ldots, B_N$（$1 \leq B_i \leq 10^5$）——每位球员的防守值。

第四行包含一个整数 $Q$（$1\leq Q\leq10^5$）——事件的数量。

接下来的 $Q$ 行中，第 $j$ 行描述第 $j$ 个事件。在任意时刻，每个 $A_i$ 和 $B_i$ 的值始终在 $1$ 到 $10^5$（含端点）之间。至少存在一个类型为 $3$ 的事件。

For each event of type $3$, output a line containing an integer representing the value of $\text{RankOverall}(k)$ at that moment.

对于每个类型为 $3$ 的事件，输出一行，包含一个整数，表示此时 $\text{RankOverall}(k)$ 的值。

At the $8$-th event, $A=[3,2,1,3,2]$ and $B=[3,6,1,4,1]$. It can be obtained that the values of $\text{RankA}$ and $\text{RankB}$ for each player are as follows:

• $\text{RankA}(1)=1$, $\text{RankB}(1)=3$
• $\text{RankA}(2)=3$, $\text{RankB}(2)=1$
• $\text{RankA}(3)=5$, $\text{RankB}(3)=4$
• $\text{RankA}(4)=1$, $\text{RankB}(4)=2$
• $\text{RankA}(5)=3$, $\text{RankB}(5)=4$

So it can be obtained that $\text{RankOverall}(1)=2$.

在第 $8$ 次活动中，$A=[3,2,1,3,2]$，$B=[3,6,1,4,1]$。可得每位选手的 $\text{RankA}$ 和 $\text{RankB}$ 值如下：

• $\text{RankA}(1)=1$，$\text{RankB}(1)=3$
• $\text{RankA}(2)=3$，$\text{RankB}(2)=1$
• $\text{RankA}(3)=5$，$\text{RankB}(3)=4$
• $\text{RankA}(4)=1$，$\text{RankB}(4)=2$
• $\text{RankA}(5)=3$，$\text{RankB}(5)=4$

因此可得 $\text{RankOverall}(1)=2$。