#include <bits/stdc++.h>
using namespace std;
struct beaver_type {
  beaver_type(unsigned long id, unsigned long ix)
      : id(id), ix(ix), prev_ix(0xFFFFFFFFLL) {}
  unsigned long id;
  unsigned long ix;
  unsigned long prev_ix;
};
int main(void) {
  ios::sync_with_stdio(false);
  unsigned long n = 0;
  cin >> n;
  vector<beaver_type> beavers;
  beavers.push_back(beaver_type(0, 0));
  vector<unsigned long> beaver_id_to_ix(n + 2);
  beaver_id_to_ix[0] = 0;
  for (unsigned long i = 1; i <= n; i++) {
    unsigned long id = 0;
    cin >> id;
    beavers.push_back(beaver_type(id, i));
    beaver_id_to_ix[id] = i;
  }
  beaver_id_to_ix[n + 1] = n + 1;
  beavers.push_back(beaver_type(n + 1, n + 1));
  for (unsigned long i = 1; i < beavers.size(); i++) {
    beaver_type& beaver = beavers[i];
    beaver.prev_ix = beaver_id_to_ix[beaver.id - 1];
  }
  unsigned long long num_queries = 0;
  cin >> num_queries;
  for (unsigned long q = 0; q < num_queries; q++) {
    unsigned long p = 0, x = 0, y = 0;
    cin >> p >> x >> y;
    if (p == 1) {
      unsigned long long count = 1;
      x++;
      for (unsigned long id = x; id <= y; id++) {
        unsigned long ix = beaver_id_to_ix[id];
        if (beavers[ix].prev_ix <= ix) continue;
        count++;
      }
      cout << count << endl;
    } else {
      beaver_type& x_place = beavers[x];
      beaver_type& y_place = beavers[y];
      beaver_type old_x = beavers[x];
      beaver_type old_y = beavers[y];
      x_place = old_y;
      y_place = old_x;
      x_place.ix = old_x.ix;
      y_place.ix = old_y.ix;
      beaver_id_to_ix[x_place.id] = x_place.ix;
      beaver_id_to_ix[y_place.id] = y_place.ix;
      beaver_type& x_post = beavers[beaver_id_to_ix[x_place.id + 1]];
      x_post.prev_ix = x_place.ix;
      beaver_type& y_post = beavers[beaver_id_to_ix[y_place.id + 1]];
      y_post.prev_ix = y_place.ix;
    }
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
struct beaver_type {
  beaver_type(unsigned long id, unsigned long ix)
      : id(id), ix(ix), prev_ix(0xFFFFFFFFLL) {}
  unsigned long id;
  unsigned long ix;
  unsigned long prev_ix;
};
int main(void) {
  ios::sync_with_stdio(false);
  unsigned long n = 0;
  cin >> n;
  vector<beaver_type> beavers;
  beavers.push_back(beaver_type(0, 0));
  vector<unsigned long> beaver_id_to_ix(n + 2);
  beaver_id_to_ix[0] = 0;
  for (unsigned long i = 1; i <= n; i++) {
    unsigned long id = 0;
    cin >> id;
    beavers.push_back(beaver_type(id, i));
    beaver_id_to_ix[id] = i;
  }
  beaver_id_to_ix[n + 1] = n + 1;
  beavers.push_back(beaver_type(n + 1, n + 1));
  for (unsigned long i = 1; i < beavers.size(); i++) {
    beaver_type& beaver = beavers[i];
    beaver.prev_ix = beaver_id_to_ix[beaver.id - 1];
  }
  unsigned long long num_queries = 0;
  cin >> num_queries;
  for (unsigned long q = 0; q < num_queries; q++) {
    unsigned long p = 0, x = 0, y = 0;
    cin >> p >> x >> y;
    if (p == 1) {
      unsigned long long count = 1;
      x++;
      for (unsigned long id = x; id <= y; id++) {
        unsigned long ix = beaver_id_to_ix[id];
        if (beavers[ix].prev_ix <= ix) continue;
        count++;
      }
      cout << count << endl;
    } else {
      beaver_type& x_place = beavers[x];
      beaver_type& y_place = beavers[y];
      beaver_type old_x = beavers[x];
      beaver_type old_y = beavers[y];
      x_place = old_y;
      y_place = old_x;
      x_place.ix = old_x.ix;
      y_place.ix = old_y.ix;
      beaver_id_to_ix[x_place.id] = x_place.ix;
      beaver_id_to_ix[y_place.id] = y_place.ix;
      beaver_type& x_post = beavers[beaver_id_to_ix[x_place.id + 1]];
      x_post.prev_ix = x_place.ix;
      beaver_type& y_post = beavers[beaver_id_to_ix[y_place.id + 1]];
      y_post.prev_ix = y_place.ix;
    }
  }
  return 0;
}

```