#include <bits/stdc++.h>
using namespace std;
const int nmax = int(3e5) + 3;
int n;
int A[nmax];
int pos[nmax];
int aib[nmax];
inline void aibUpdate(int pos, int val) {
  for (; pos <= n; pos += (pos & -pos)) {
    aib[pos] += val;
  }
}
inline int aibQuery(int pos) {
  int ret = 0;
  for (; pos > 0; pos -= (pos & -pos)) {
    ret += aib[pos];
  }
  return ret;
}
inline void check(int a) {
  if (a > n) return;
  if (pos[a] > pos[a - 1]) {
    if (aibQuery(a) - aibQuery(a - 1) > 0) {
      aibUpdate(a, -1);
    }
  } else {
    if (aibQuery(a) - aibQuery(a - 1) == 0) {
      aibUpdate(a, 1);
    }
  }
}
inline void update(int a, int b) {
  swap(A[a], A[b]);
  swap(pos[A[a]], pos[A[b]]);
  check(A[a]);
  check(A[a] + 1);
  check(A[b]);
  check(A[b] + 1);
}
inline int query(int a, int b) { return aibQuery(b) - aibQuery(a) + 1; }
inline int queryBrute(int a, int b) {
  int ret = 1;
  for (int i = a + 1; i <= b; i++) {
    if (pos[i] < pos[i - 1]) {
      ret++;
    }
  }
  return ret;
}
void generateData() {
  srand(time(0));
  cout << "100\n";
  vector<int> v(100);
  for (int i = 1; i <= 100; i++) {
    v[i - 1] = i;
  }
  while (v.empty() == false) {
    int p;
    cout << v[p = rand() % v.size()] << " ";
    v.erase(v.begin() + p);
  }
  cout << "\n";
  cout << 1000 << "\n";
  for (int i = 1; i <= 1000; i++) {
    int val = rand() % 100 + 1;
    cout << rand() % 2 + 1 << " " << val << " ";
    cout << rand() % (100 - val + 1) + val << "\n";
  }
  exit(0);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    pos[A[i]] = i;
  }
  for (int i = 2; i <= n; i++) {
    if (pos[i] < pos[i - 1]) {
      aibUpdate(i, 1);
    }
  }
  int q, t, a, b;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> t >> a >> b;
    if (t == 1) {
      cout << query(a, b) << "\n";
    } else {
      update(a, b);
    }
  }
  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;
const int nmax = int(3e5) + 3;
int n;
int A[nmax];
int pos[nmax];
int aib[nmax];
inline void aibUpdate(int pos, int val) {
  for (; pos <= n; pos += (pos & -pos)) {
    aib[pos] += val;
  }
}
inline int aibQuery(int pos) {
  int ret = 0;
  for (; pos > 0; pos -= (pos & -pos)) {
    ret += aib[pos];
  }
  return ret;
}
inline void check(int a) {
  if (a > n) return;
  if (pos[a] > pos[a - 1]) {
    if (aibQuery(a) - aibQuery(a - 1) > 0) {
      aibUpdate(a, -1);
    }
  } else {
    if (aibQuery(a) - aibQuery(a - 1) == 0) {
      aibUpdate(a, 1);
    }
  }
}
inline void update(int a, int b) {
  swap(A[a], A[b]);
  swap(pos[A[a]], pos[A[b]]);
  check(A[a]);
  check(A[a] + 1);
  check(A[b]);
  check(A[b] + 1);
}
inline int query(int a, int b) { return aibQuery(b) - aibQuery(a) + 1; }
inline int queryBrute(int a, int b) {
  int ret = 1;
  for (int i = a + 1; i <= b; i++) {
    if (pos[i] < pos[i - 1]) {
      ret++;
    }
  }
  return ret;
}
void generateData() {
  srand(time(0));
  cout << "100\n";
  vector<int> v(100);
  for (int i = 1; i <= 100; i++) {
    v[i - 1] = i;
  }
  while (v.empty() == false) {
    int p;
    cout << v[p = rand() % v.size()] << " ";
    v.erase(v.begin() + p);
  }
  cout << "\n";
  cout << 1000 << "\n";
  for (int i = 1; i <= 1000; i++) {
    int val = rand() % 100 + 1;
    cout << rand() % 2 + 1 << " " << val << " ";
    cout << rand() % (100 - val + 1) + val << "\n";
  }
  exit(0);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    pos[A[i]] = i;
  }
  for (int i = 2; i <= n; i++) {
    if (pos[i] < pos[i - 1]) {
      aibUpdate(i, 1);
    }
  }
  int q, t, a, b;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> t >> a >> b;
    if (t == 1) {
      cout << query(a, b) << "\n";
    } else {
      update(a, b);
    }
  }
  return 0;
}

```