#include <bits/stdc++.h>
using namespace std;
int n, m, a[300002], sv[300002], u, x, y;
struct fenw {
  int x[300002], tag[300002];
  void update(int now, int val) {
    if (val == 1 && tag[now] == 0) {
      tag[now] = 1;
      for (int i = now; i <= n; i += (i & (-i))) x[i] += val;
    } else if (val == -1 && tag[now] == 1) {
      tag[now] = 0;
      for (int i = now; i <= n; i += (i & (-i))) x[i] += val;
    }
  }
  int find(int now) {
    int s = 0;
    for (int i = now; i > 0; i -= (i & (-i))) s += x[i];
    return s;
  }
  void pr() {
    for (int i = 1; i <= n; i++) printf("%d ", x[i]);
    printf("\n");
    for (int i = 1; i <= n; i++) printf("%d ", tag[i]);
    printf("\n");
  }
};
fenw f;
int main() {
  scanf("%d", &n);
  sv[n + 1] = n + 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    sv[a[i]] = i;
  }
  for (int i = 1; i <= n; i++)
    if (sv[i] < sv[i - 1]) f.update(i - 1, 1);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &u, &x, &y);
    if (u == 1)
      printf("%d\n", f.find(y - 1) - f.find(x - 1) + 1);
    else {
      swap(sv[a[x]], sv[a[y]]);
      u = sv[a[x] - 1] > sv[a[x]] ? 1 : -1;
      f.update(a[x] - 1, u);
      u = sv[a[x]] > sv[a[x] + 1] ? 1 : -1;
      f.update(a[x], u);
      u = sv[a[y]] > sv[a[y] + 1] ? 1 : -1;
      f.update(a[y], u);
      if (x != y - 1) {
        u = sv[a[y] - 1] > sv[a[y]] ? 1 : -1;
        f.update(a[y] - 1, u);
      }
      swap(a[x], a[y]);
    }
  }
  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;
int n, m, a[300002], sv[300002], u, x, y;
struct fenw {
  int x[300002], tag[300002];
  void update(int now, int val) {
    if (val == 1 && tag[now] == 0) {
      tag[now] = 1;
      for (int i = now; i <= n; i += (i & (-i))) x[i] += val;
    } else if (val == -1 && tag[now] == 1) {
      tag[now] = 0;
      for (int i = now; i <= n; i += (i & (-i))) x[i] += val;
    }
  }
  int find(int now) {
    int s = 0;
    for (int i = now; i > 0; i -= (i & (-i))) s += x[i];
    return s;
  }
  void pr() {
    for (int i = 1; i <= n; i++) printf("%d ", x[i]);
    printf("\n");
    for (int i = 1; i <= n; i++) printf("%d ", tag[i]);
    printf("\n");
  }
};
fenw f;
int main() {
  scanf("%d", &n);
  sv[n + 1] = n + 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    sv[a[i]] = i;
  }
  for (int i = 1; i <= n; i++)
    if (sv[i] < sv[i - 1]) f.update(i - 1, 1);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &u, &x, &y);
    if (u == 1)
      printf("%d\n", f.find(y - 1) - f.find(x - 1) + 1);
    else {
      swap(sv[a[x]], sv[a[y]]);
      u = sv[a[x] - 1] > sv[a[x]] ? 1 : -1;
      f.update(a[x] - 1, u);
      u = sv[a[x]] > sv[a[x] + 1] ? 1 : -1;
      f.update(a[x], u);
      u = sv[a[y]] > sv[a[y] + 1] ? 1 : -1;
      f.update(a[y], u);
      if (x != y - 1) {
        u = sv[a[y] - 1] > sv[a[y]] ? 1 : -1;
        f.update(a[y] - 1, u);
      }
      swap(a[x], a[y]);
    }
  }
  return 0;
}

```