#include <bits/stdc++.h>
using namespace std;
const int MaxN = 5000005;
typedef int Tar[MaxN];
int n, loai, x, y, nques;
Tar a, b, val, l, h;
void Build(int x, int low, int high) {
  l[x] = low;
  h[x] = high;
  val[x] = 0;
  if (low != high) {
    int mid = (low + high) / 2;
    Build(2 * x, low, mid);
    Build(2 * x + 1, mid + 1, high);
  }
}
void Update(int i, int x, int value) {
  if (l[i] == h[i])
    val[i] = value;
  else {
    if (x <= h[2 * i])
      Update(i * 2, x, value);
    else
      Update(i * 2 + 1, x, value);
    val[i] = val[i * 2] + val[i * 2 + 1];
  }
}
int Get(int i, int low, int high) {
  if ((low <= l[i]) && (h[i] <= high))
    return val[i];
  else {
    int ans = 0;
    if (low <= h[2 * i]) ans += Get(i * 2, low, high);
    if (high > h[2 * i]) ans += Get(i * 2 + 1, low, high);
    return ans;
  }
}
void init() {
  Build(1, 1, n);
  for (int i = 1; i <= n - 1; i++)
    if (b[i] < b[i + 1]) {
      Update(1, i, 0);
    } else
      Update(1, i, 1);
}
void Swap(int x, int y) {
  int tmp;
  tmp = a[x];
  a[x] = a[y];
  a[y] = tmp;
  b[a[x]] = x;
  b[a[y]] = y;
  if (a[x] != 1) {
    if (b[a[x] - 1] < b[a[x]])
      Update(1, a[x] - 1, 0);
    else
      Update(1, a[x] - 1, 1);
  }
  if (a[x] != n) {
    if (b[a[x]] < b[a[x] + 1])
      Update(1, a[x], 0);
    else
      Update(1, a[x], 1);
  }
  if (a[y] != 1) {
    if (b[a[y] - 1] < b[a[y]])
      Update(1, a[y] - 1, 0);
    else
      Update(1, a[y] - 1, 1);
  }
  if (a[y] != n) {
    if (b[a[y]] < b[a[y] + 1])
      Update(1, a[y], 0);
    else
      Update(1, a[y], 1);
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[a[i]] = i;
  }
  init();
  cin >> nques;
  for (int ques = 1; ques <= nques; ques++) {
    cin >> loai >> x >> y;
    if (loai == 2)
      Swap(x, y);
    else
      cout << Get(1, x, y - 1) + 1 << endl;
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 MaxN = 5000005;
typedef int Tar[MaxN];
int n, loai, x, y, nques;
Tar a, b, val, l, h;
void Build(int x, int low, int high) {
  l[x] = low;
  h[x] = high;
  val[x] = 0;
  if (low != high) {
    int mid = (low + high) / 2;
    Build(2 * x, low, mid);
    Build(2 * x + 1, mid + 1, high);
  }
}
void Update(int i, int x, int value) {
  if (l[i] == h[i])
    val[i] = value;
  else {
    if (x <= h[2 * i])
      Update(i * 2, x, value);
    else
      Update(i * 2 + 1, x, value);
    val[i] = val[i * 2] + val[i * 2 + 1];
  }
}
int Get(int i, int low, int high) {
  if ((low <= l[i]) && (h[i] <= high))
    return val[i];
  else {
    int ans = 0;
    if (low <= h[2 * i]) ans += Get(i * 2, low, high);
    if (high > h[2 * i]) ans += Get(i * 2 + 1, low, high);
    return ans;
  }
}
void init() {
  Build(1, 1, n);
  for (int i = 1; i <= n - 1; i++)
    if (b[i] < b[i + 1]) {
      Update(1, i, 0);
    } else
      Update(1, i, 1);
}
void Swap(int x, int y) {
  int tmp;
  tmp = a[x];
  a[x] = a[y];
  a[y] = tmp;
  b[a[x]] = x;
  b[a[y]] = y;
  if (a[x] != 1) {
    if (b[a[x] - 1] < b[a[x]])
      Update(1, a[x] - 1, 0);
    else
      Update(1, a[x] - 1, 1);
  }
  if (a[x] != n) {
    if (b[a[x]] < b[a[x] + 1])
      Update(1, a[x], 0);
    else
      Update(1, a[x], 1);
  }
  if (a[y] != 1) {
    if (b[a[y] - 1] < b[a[y]])
      Update(1, a[y] - 1, 0);
    else
      Update(1, a[y] - 1, 1);
  }
  if (a[y] != n) {
    if (b[a[y]] < b[a[y] + 1])
      Update(1, a[y], 0);
    else
      Update(1, a[y], 1);
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[a[i]] = i;
  }
  init();
  cin >> nques;
  for (int ques = 1; ques <= nques; ques++) {
    cin >> loai >> x >> y;
    if (loai == 2)
      Swap(x, y);
    else
      cout << Get(1, x, y - 1) + 1 << endl;
  }
  return 0;
}

```