#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ")";
}
string to_string(string s) { return s; }
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
long long n, q;
vector<long long> p;
vector<long long> ending;
vector<long long> where;
void update_ending(long long e) {
  if (e + 1 < n) ending[e] = (where[e + 1] < where[e] ? 1LL : 0LL);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  p.resize(n);
  where.resize(n);
  for (long long i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
    where[p[i]] = i;
  }
  ending.resize(n);
  for (long long i = 0; i < n - 1; i++) {
    update_ending(i);
  }
  long long q;
  cin >> q;
  while (q--) {
    long long e, x, y;
    cin >> e >> x >> y;
    x--;
    y--;
    if (e == 1) {
      long long cnt = 0;
      for (long long i = x; i < y; i++) {
        cnt += ending[i];
      }
      cout << cnt + 1 << '\n';
    } else {
      long long nx = p[x], ny = p[y];
      swap(p[x], p[y]);
      swap(where[nx], where[ny]);
      if (nx > 0) update_ending(nx - 1);
      if (nx < n) update_ending(nx);
      if (ny > 0) update_ending(ny - 1);
      if (ny < n) update_ending(ny);
    }
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ")";
}
string to_string(string s) { return s; }
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
long long n, q;
vector<long long> p;
vector<long long> ending;
vector<long long> where;
void update_ending(long long e) {
  if (e + 1 < n) ending[e] = (where[e + 1] < where[e] ? 1LL : 0LL);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  p.resize(n);
  where.resize(n);
  for (long long i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
    where[p[i]] = i;
  }
  ending.resize(n);
  for (long long i = 0; i < n - 1; i++) {
    update_ending(i);
  }
  long long q;
  cin >> q;
  while (q--) {
    long long e, x, y;
    cin >> e >> x >> y;
    x--;
    y--;
    if (e == 1) {
      long long cnt = 0;
      for (long long i = x; i < y; i++) {
        cnt += ending[i];
      }
      cout << cnt + 1 << '\n';
    } else {
      long long nx = p[x], ny = p[y];
      swap(p[x], p[y]);
      swap(where[nx], where[ny]);
      if (nx > 0) update_ending(nx - 1);
      if (nx < n) update_ending(nx);
      if (ny > 0) update_ending(ny - 1);
      if (ny < n) update_ending(ny);
    }
  }
  return 0;
}

```