#include <bits/stdc++.h>
using namespace std;
int dx4[] = {-1, 0, 1, 0};
int dy4[] = {0, 1, 0, -1};
int dx8[] = {-1, -1, 0, 1, 1, 1, 0, -1};
int dy8[] = {0, 1, 1, 1, 0, -1, -1, -1};
bool AreEqual(double a, double b) { return (fabs(a - b) < 1e-10); }
template <class T>
bool GetNr(T &_value) {
  T _sign = 1;
  char ch;
  _value = 0;
  while (!isdigit(ch = getchar())) {
    if (ch == -1) return false;
    ch == '-' ? _sign = -1 : _sign = 1;
  }
  do {
    _value = _value * 10 + (ch - '0');
  } while (isdigit(ch = getchar()));
  _value *= _sign;
  return true;
}
int N, Q, cnt = 1;
vector<int> values, positions, aib, checked;
void InsertAib(int pos, int val) {
  if (checked[pos] == cnt) return;
  checked[pos] = cnt;
  for (; pos <= N; pos += pos & (-pos)) {
    aib[pos] += val;
  }
}
int QueryAib(int pos) {
  int ans = 0;
  for (; pos; pos -= pos & (-pos)) {
    ans += aib[pos];
  }
  return ans;
}
void Verif(int val, int add) {
  if (val > 1 && positions[val - 1] > positions[val]) InsertAib(val - 1, add);
  if (val < N && positions[val] > positions[val + 1]) InsertAib(val, add);
}
int main() {
  int i;
  GetNr(N);
  values.resize(1 + N);
  positions.resize(1 + N);
  checked.resize(1 + N);
  aib.resize(1 + N);
  for (i = 1; i <= N; ++i) {
    GetNr(values[i]);
    positions[values[i]] = i;
  }
  for (i = 1; i <= N; ++i) {
    if (positions[i + 1] < positions[i]) {
      InsertAib(i, 1);
    }
  }
  int t, x, y;
  GetNr(Q);
  for (i = 0; i < Q; ++i) {
    GetNr(t);
    GetNr(x);
    GetNr(y);
    if (t == 1) {
      printf("%d\n", 1 + QueryAib(y - 1) - QueryAib(x - 1));
    } else {
      x = values[x];
      y = values[y];
      ++cnt;
      Verif(x, -1);
      Verif(y, -1);
      swap(values[positions[x]], values[positions[y]]);
      swap(positions[x], positions[y]);
      ++cnt;
      Verif(x, 1);
      Verif(y, 1);
    }
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution 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;
int dx4[] = {-1, 0, 1, 0};
int dy4[] = {0, 1, 0, -1};
int dx8[] = {-1, -1, 0, 1, 1, 1, 0, -1};
int dy8[] = {0, 1, 1, 1, 0, -1, -1, -1};
bool AreEqual(double a, double b) { return (fabs(a - b) < 1e-10); }
template <class T>
bool GetNr(T &_value) {
  T _sign = 1;
  char ch;
  _value = 0;
  while (!isdigit(ch = getchar())) {
    if (ch == -1) return false;
    ch == '-' ? _sign = -1 : _sign = 1;
  }
  do {
    _value = _value * 10 + (ch - '0');
  } while (isdigit(ch = getchar()));
  _value *= _sign;
  return true;
}
int N, Q, cnt = 1;
vector<int> values, positions, aib, checked;
void InsertAib(int pos, int val) {
  if (checked[pos] == cnt) return;
  checked[pos] = cnt;
  for (; pos <= N; pos += pos & (-pos)) {
    aib[pos] += val;
  }
}
int QueryAib(int pos) {
  int ans = 0;
  for (; pos; pos -= pos & (-pos)) {
    ans += aib[pos];
  }
  return ans;
}
void Verif(int val, int add) {
  if (val > 1 && positions[val - 1] > positions[val]) InsertAib(val - 1, add);
  if (val < N && positions[val] > positions[val + 1]) InsertAib(val, add);
}
int main() {
  int i;
  GetNr(N);
  values.resize(1 + N);
  positions.resize(1 + N);
  checked.resize(1 + N);
  aib.resize(1 + N);
  for (i = 1; i <= N; ++i) {
    GetNr(values[i]);
    positions[values[i]] = i;
  }
  for (i = 1; i <= N; ++i) {
    if (positions[i + 1] < positions[i]) {
      InsertAib(i, 1);
    }
  }
  int t, x, y;
  GetNr(Q);
  for (i = 0; i < Q; ++i) {
    GetNr(t);
    GetNr(x);
    GetNr(y);
    if (t == 1) {
      printf("%d\n", 1 + QueryAib(y - 1) - QueryAib(x - 1));
    } else {
      x = values[x];
      y = values[y];
      ++cnt;
      Verif(x, -1);
      Verif(y, -1);
      swap(values[positions[x]], values[positions[y]]);
      swap(positions[x], positions[y]);
      ++cnt;
      Verif(x, 1);
      Verif(y, 1);
    }
  }
  return 0;
}

```