#include <bits/stdc++.h>
using namespace std;
template <class A, class B>
ostream& operator<<(ostream& o, pair<A, B>& p) {
  return o << "(" << p.first << ", " << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "{";
  for (typeof((v).begin()) it = (v).begin(); it != (v).end(); it++)
    o << *it << ", ";
  return o << "}";
}
struct debugger {
  ostream& out;
  bool first;
  void start() { first = true; }
  debugger(ostream& out) : out(out) { start(); }
  void nl() {
    out << endl;
    start();
  }
  template <class T>
  debugger& operator,(T& v) {
    if (first)
      first = false;
    else
      out << ", ";
    out << v;
    return *this;
  }
  template <class T>
  debugger& operator,(pair<T*, int> arr) {
    out << "{";
    for (typeof((arr.second) - 1) i = (0); i <= ((arr.second) - 1); i++) {
      out << arr.first[i] << ", ";
    }
    out << "}";
    return *this;
  }
} dbg(cerr);
int tree[1 << 18];
int MaxVal = 1 << 17;
int read(int idx) {
  int sum = 0;
  while (idx > 0) {
    sum += tree[idx];
    idx -= (idx & -idx);
  }
  return sum;
}
void update(int idx, int val) {
  while (idx <= MaxVal) {
    tree[idx] += val;
    idx += (idx & -idx);
  }
}
int arr[111111], p[111111], inv[111111];
void f(int b) {
  if (p[arr[b]] < p[arr[b] - 1] && inv[arr[b] - 1] == 0) {
    inv[arr[b] - 1] = 1;
    update(arr[b] - 1, 1);
  }
  if (p[arr[b]] > p[arr[b] - 1] && inv[arr[b] - 1] == 1) {
    inv[arr[b] - 1] = 0;
    update(arr[b] - 1, -1);
  }
  if (p[arr[b]] > p[arr[b] + 1] && inv[arr[b]] == 0) {
    inv[arr[b]] = 1;
    update(arr[b], 1);
  }
  if (p[arr[b]] < p[arr[b] + 1] && inv[arr[b]] == 1) {
    inv[arr[b]] = 0;
    update(arr[b], -1);
  }
}
int main() {
  int n, q, a, b, t;
  cin >> n;
  for (typeof(n) i = (1); i <= (n); i++) {
    cin >> arr[i];
    p[arr[i]] = i;
  }
  for (typeof(n - 1) i = (1); i <= (n - 1); i++) {
    if (p[i + 1] < p[i]) {
      inv[i] = 1;
      update(i, 1);
    }
  }
  cin >> q;
  for (typeof((q)-1) i = (0); i <= ((q)-1); i++) {
    cin >> t;
    if (t == 1) {
      cin >> a >> b;
      cout << read(b - 1) - read(a - 1) + 1 << endl;
    } else {
      cin >> a >> b;
      swap(p[arr[b]], p[arr[a]]);
      f(b);
      f(a);
      swap(arr[b], arr[a]);
    }
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 <class A, class B>
ostream& operator<<(ostream& o, pair<A, B>& p) {
  return o << "(" << p.first << ", " << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "{";
  for (typeof((v).begin()) it = (v).begin(); it != (v).end(); it++)
    o << *it << ", ";
  return o << "}";
}
struct debugger {
  ostream& out;
  bool first;
  void start() { first = true; }
  debugger(ostream& out) : out(out) { start(); }
  void nl() {
    out << endl;
    start();
  }
  template <class T>
  debugger& operator,(T& v) {
    if (first)
      first = false;
    else
      out << ", ";
    out << v;
    return *this;
  }
  template <class T>
  debugger& operator,(pair<T*, int> arr) {
    out << "{";
    for (typeof((arr.second) - 1) i = (0); i <= ((arr.second) - 1); i++) {
      out << arr.first[i] << ", ";
    }
    out << "}";
    return *this;
  }
} dbg(cerr);
int tree[1 << 18];
int MaxVal = 1 << 17;
int read(int idx) {
  int sum = 0;
  while (idx > 0) {
    sum += tree[idx];
    idx -= (idx & -idx);
  }
  return sum;
}
void update(int idx, int val) {
  while (idx <= MaxVal) {
    tree[idx] += val;
    idx += (idx & -idx);
  }
}
int arr[111111], p[111111], inv[111111];
void f(int b) {
  if (p[arr[b]] < p[arr[b] - 1] && inv[arr[b] - 1] == 0) {
    inv[arr[b] - 1] = 1;
    update(arr[b] - 1, 1);
  }
  if (p[arr[b]] > p[arr[b] - 1] && inv[arr[b] - 1] == 1) {
    inv[arr[b] - 1] = 0;
    update(arr[b] - 1, -1);
  }
  if (p[arr[b]] > p[arr[b] + 1] && inv[arr[b]] == 0) {
    inv[arr[b]] = 1;
    update(arr[b], 1);
  }
  if (p[arr[b]] < p[arr[b] + 1] && inv[arr[b]] == 1) {
    inv[arr[b]] = 0;
    update(arr[b], -1);
  }
}
int main() {
  int n, q, a, b, t;
  cin >> n;
  for (typeof(n) i = (1); i <= (n); i++) {
    cin >> arr[i];
    p[arr[i]] = i;
  }
  for (typeof(n - 1) i = (1); i <= (n - 1); i++) {
    if (p[i + 1] < p[i]) {
      inv[i] = 1;
      update(i, 1);
    }
  }
  cin >> q;
  for (typeof((q)-1) i = (0); i <= ((q)-1); i++) {
    cin >> t;
    if (t == 1) {
      cin >> a >> b;
      cout << read(b - 1) - read(a - 1) + 1 << endl;
    } else {
      cin >> a >> b;
      swap(p[arr[b]], p[arr[a]]);
      f(b);
      f(a);
      swap(arr[b], arr[a]);
    }
  }
  return 0;
}

```