Codeforces Round #416 (Div. 2) has been moved to start on 27.05.2017 09:05 (UTC).
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B2. Shave Beaver!

time limit per test

2 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputThe 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.

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* ≤ 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.

Output

For each query with *p*_{i} = 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

Codeforces (c) Copyright 2010-2017 Mike Mirzayanov

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