A9133. Shave Beaver!

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 $a_{1},a_{2},...,a_{n}$ 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 $i_{1}<i_{2}<...<i_{k}$ , that $a_{i1}=x$ , $a_{i2}=x+1$ , ..., $a_{ik-1}=y-1$ , $a_{ik}=y$ . 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 $[x,p_{1}]$ , $[p_{1}+1,p_{2}]$ , ..., $[p_{m}+1,y]$ $(x<=p_{1}<p_{2}<...<p_{m}<y)$ , 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 $a_{x}$ and $a_{y}$ ) swapped.

You can assume that any beaver can be shaved any number of times.

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, $1<=q<=10^{5}$ . The next $q$ lines contain the queries. Each query $i$ looks as $p_{i}$ $x_{i}$ $y_{i}$ , where $p_{i}$ is the query type ( $1$ is to shave beavers from $x_{i}$ to $y_{i}$ , inclusive, $2$ is to swap beavers on positions $x_{i}$ and $y_{i}$ ). All queries meet the condition: $1<=x_{i}<y_{i}<=n$ .

- 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: $n<=3·10^{5}$ (subproblems B1+B2).

Note that the number of queries $q$ is limited $1<=q<=10^{5}$ in both subproblem B1 and subproblem B2.

For each query with $p_{i}=1$ , print the minimum number of Beavershave 5000 sessions.