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6013: Shave Beaver!

内存限制：256 MB 时间限制：1 S
题面：传统 评测方式：文本比较 上传者：
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题目描述

Smart Beaver最近设计并制造了一款创新的纳米技术通用海狸质量剃须机“Beavershave 5000”。海狸剃须5000可以家庭剃海狸！它是如何工作的？非常容易！
有 n 只海狸，每只海狸都有一个从 1 到 n 的唯一 id。考虑排列一个a1，a2,...,an这些海狸。Beavershave 5000 需要一段来剃除 ID 从 x 到 y（含）的海狸，当且仅当存在此类索引时i1<i2<...<ik。ai1=x,ai2=x+1, ...,aiK-1=Y-1,aik=y。这真的很方便。例如，它需要一段来剃掉海狸 1，2，3,...,n 的排列。
如果我们不能在一个会话中将海狸从 x 剃到 y，那么我们可以将这些海狸分成几组[x，p1],[p1+1，p2], ...,[pm+1，y]

] (x≤p1<p2<...<pm<和)，这样机器就可以在一个会话中剃掉每组中的海狸。但是，Beavershave 5000 需要 m+1 个段才能将海狸从 x 剃到 y。
所有的海狸都焦躁不安，它们一直试图交换。因此，如果我们更正式地考虑问题，我们可以考虑两种类型的查询：

· Beavershave 5000 需要多少次来剃除 ID 从 X 到 Y 的海狸（包含）？

· 位置 X 和 Y 上的两只海狸（海狸ax和ay） 交换。

您可以假设任何海狸都可以剃任意次数.

The Smart Beaver has recently designed and built an innovative nanotechnologic all-purpose beaver mass shaving machine, "Beavershave 5000". Beavershave 5000 can shave beavers by families! How does it work? Very easily!

There are n beavers, each of them has a unique id from 1 to n. Consider a permutation a1,a2,...,an of n these beavers. Beavershave 5000 needs one session to shave beavers with ids from x to y (inclusive) if and only if there are such indices i1<i2<...<ik, that ai1=x, ai2=x+1, ..., aik-1=y-1, aik=y. And that is really convenient. For example, it needs one session to shave a permutation of beavers 1,2,3,...,n.
If we can't shave beavers from x to y in one session, then we can split these beavers into groups [x,p1], [p1+1,p2], ..., [pm+1,y] (x≤p1<p2<...<pm<y), in such a way that the machine can shave beavers in each group in one session. But then Beavershave 5000 needs m+1 working sessions to shave beavers from x to y.
All beavers are restless and they keep trying to swap. So if we consider the problem more formally, we can consider queries of two types:
  • what is the minimum number of sessions that Beavershave 5000 needs to shave beavers with ids from x to y, inclusive?
  • two beavers on positions x and y (the beavers ax and ay) swapped.
You can assume that any beaver can be shaved any number of times.
Input
The first line contains integer n − the total number of beavers, 2≤n. The second line contains n space-separated integers − the initial beaver permutation.
The third line contains integer q − the number of queries, 1≤q≤105. The next q lines contain the queries. Each query i looks as pi xi yi, where pi is the query type (1 is to shave beavers from xi to yi, inclusive, 2 is to swap beavers on positions xi and yi). All queries meet the condition: 1≤xi<yi≤n.
  • to get 30 points, you need to solve the problem with constraints: n≤100 (subproblem B1);
  • to get 100 points, you need to solve the problem with constraints: n≤3·105 (subproblems B1+B2).
Note that the number of queries q is limited 1≤q≤105 in both subproblem B1 and subproblem B2.
Output
For each query with pi=1, print the minimum number of Beavershave 5000 sessions.
Examples
Input 
5
1 3 4 2 5
6
1 1 5
1 3 4
2 2 3
1 1 5
2 1 5
1 1 5
Output 
2
1
3
5

输入格式

第一行包含整数 n − 海狸总数 2≤n。第二行包含 n 个空格分隔的整数 - 初始海狸排列。
第三行包含整数 q − 查询数， 1≤q≤105..接下来的 q 行包含查询。每个查询1<=q<=105。查询类型（1 是从中剃掉海狸xi到yi，包括，2是交换海狸的位置xi和yi).所有查询都满足条件：1≤xi<yi≤n.

· 要获得 30 分，您需要使用约束求解问题：n≤100（子问题 B1）;

· 要获得 100 分，您需要使用约束来解决问题：n≤3·105（子问题 B1+B2）。

请注意，查询次数 q 是有限的1≤q≤105在子问题 B1 和子问题 B2 中。

输出格式

对于每个查询，使用pi=1，打印海狸剃须 5000 的最小数量。
Input 

5
1 3 4 2 5
6
1 1 5
1 3 4
2 2 3
1 1 5
2 1 5
1 1 5
Output 
2
1
3
5

输入样例 复制 

5
1 3 4 2 5
6
1 1 5
1 3 4
2 2 3
1 1 5
2 1 5
1 1 5

输出样例 复制 

2
1
3
5

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