#include <bits/stdc++.h>
using namespace std;
int n, nevents;
int p[300000];
int inv[300000];
int val[1 << 19];
int get(int x) {
  ++x;
  int ret = 0;
  while (x != 0) {
    ret += val[x];
    x -= x & -x;
  }
  return ret;
}
void update(int x, int by) {
  ++x;
  while (x != 1 << 19) {
    val[x] += by;
    x += x & -x;
  }
}
void run(int casenr) {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) {
    scanf("%d", &p[i]);
    --p[i];
  }
  for (int i = (0); i < (n); ++i) inv[p[i]] = i;
  memset(val, 0, sizeof(val));
  for (int i = (0); i < (n - 1); ++i)
    if (inv[i + 1] < inv[i]) update(i, +1);
  scanf("%d", &nevents);
  for (int e = (0); e < (nevents); ++e) {
    int type, x, y;
    scanf("%d%d%d", &type, &x, &y);
    --x, --y;
    if (type == 1) {
      int ans = get(y - 1) - get(x - 1) + 1;
      printf("%d\n", ans);
    }
    if (type == 2) {
      int a = p[x], b = p[y];
      int lst[4], nlst = 0;
      if (a - 1 >= 0) lst[nlst++] = a - 1;
      if (a + 1 < n) lst[nlst++] = a;
      if (b - 1 >= 0) lst[nlst++] = b - 1;
      if (b + 1 < n) lst[nlst++] = b;
      sort(lst, lst + nlst);
      for (int i = (0); i < (nlst); ++i)
        if (i == 0 || lst[i] != lst[i - 1]) {
          int xi = lst[i], xj = xi + 1, pi = inv[xi], pj = inv[xj],
              ni = pi == x   ? y
                   : pi == y ? x
                             : pi,
              nj = pj == x   ? y
                   : pj == y ? x
                             : pj;
          int delta = (nj < ni ? 1 : 0) - (pj < pi ? 1 : 0);
          if (delta != 0) update(xi, delta);
        }
      swap(p[x], p[y]);
      inv[p[x]] = x;
      inv[p[y]] = y;
    }
  }
}
int main() {
  run(1);
  return 0;
}

### Prompt

Generate 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 n, nevents;
int p[300000];
int inv[300000];
int val[1 << 19];
int get(int x) {
  ++x;
  int ret = 0;
  while (x != 0) {
    ret += val[x];
    x -= x & -x;
  }
  return ret;
}
void update(int x, int by) {
  ++x;
  while (x != 1 << 19) {
    val[x] += by;
    x += x & -x;
  }
}
void run(int casenr) {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) {
    scanf("%d", &p[i]);
    --p[i];
  }
  for (int i = (0); i < (n); ++i) inv[p[i]] = i;
  memset(val, 0, sizeof(val));
  for (int i = (0); i < (n - 1); ++i)
    if (inv[i + 1] < inv[i]) update(i, +1);
  scanf("%d", &nevents);
  for (int e = (0); e < (nevents); ++e) {
    int type, x, y;
    scanf("%d%d%d", &type, &x, &y);
    --x, --y;
    if (type == 1) {
      int ans = get(y - 1) - get(x - 1) + 1;
      printf("%d\n", ans);
    }
    if (type == 2) {
      int a = p[x], b = p[y];
      int lst[4], nlst = 0;
      if (a - 1 >= 0) lst[nlst++] = a - 1;
      if (a + 1 < n) lst[nlst++] = a;
      if (b - 1 >= 0) lst[nlst++] = b - 1;
      if (b + 1 < n) lst[nlst++] = b;
      sort(lst, lst + nlst);
      for (int i = (0); i < (nlst); ++i)
        if (i == 0 || lst[i] != lst[i - 1]) {
          int xi = lst[i], xj = xi + 1, pi = inv[xi], pj = inv[xj],
              ni = pi == x   ? y
                   : pi == y ? x
                             : pi,
              nj = pj == x   ? y
                   : pj == y ? x
                             : pj;
          int delta = (nj < ni ? 1 : 0) - (pj < pi ? 1 : 0);
          if (delta != 0) update(xi, delta);
        }
      swap(p[x], p[y]);
      inv[p[x]] = x;
      inv[p[y]] = y;
    }
  }
}
int main() {
  run(1);
  return 0;
}

```