#include <bits/stdc++.h>
using namespace std;
int n, q, a[300005], r[300005];
struct bit {
  int tree[300005];
  void add(int x, int v) {
    while (x <= n) {
      tree[x] += v;
      x += x & -x;
    }
  }
  int sum(int x) {
    int ret = 0;
    while (x) {
      ret += tree[x];
      x -= x & -x;
    }
    return ret;
  }
} bit;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    r[a[i]] = i;
  }
  for (int i = 1; i < n; i++) {
    if (r[i] > r[i + 1]) bit.add(i, 1);
  }
  scanf("%d", &q);
  while (q--) {
    int p, x, y;
    scanf("%d %d %d", &p, &x, &y);
    if (p == 1) {
      printf("%d\n", bit.sum(y - 1) - bit.sum(x - 1) + 1);
    } else {
      set<int> s;
      if (r[a[x] - 1] > r[a[x]] && a[x] > 1) s.insert(a[x] - 1);
      if (r[a[x]] > r[a[x] + 1] && a[x] + 1 <= n) s.insert(a[x]);
      if (r[a[y] - 1] > r[a[y]] && a[y] > 1) s.insert(a[y] - 1);
      if (r[a[y]] > r[a[y] + 1] && a[y] + 1 <= n) s.insert(a[y]);
      for (auto &i : s) {
        bit.add(i, -1);
      }
      s.clear();
      swap(a[x], a[y]);
      swap(r[a[x]], r[a[y]]);
      if (r[a[x] - 1] > r[a[x]] && a[x] > 1) s.insert(a[x] - 1);
      if (r[a[x]] > r[a[x] + 1] && a[x] + 1 <= n) s.insert(a[x]);
      if (r[a[y] - 1] > r[a[y]] && a[y] > 1) s.insert(a[y] - 1);
      if (r[a[y]] > r[a[y] + 1] && a[y] + 1 <= n) s.insert(a[y]);
      for (auto &i : s) {
        bit.add(i, 1);
      }
      s.clear();
    }
  }
}

### Prompt

In Cpp, your task is to solve 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, q, a[300005], r[300005];
struct bit {
  int tree[300005];
  void add(int x, int v) {
    while (x <= n) {
      tree[x] += v;
      x += x & -x;
    }
  }
  int sum(int x) {
    int ret = 0;
    while (x) {
      ret += tree[x];
      x -= x & -x;
    }
    return ret;
  }
} bit;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    r[a[i]] = i;
  }
  for (int i = 1; i < n; i++) {
    if (r[i] > r[i + 1]) bit.add(i, 1);
  }
  scanf("%d", &q);
  while (q--) {
    int p, x, y;
    scanf("%d %d %d", &p, &x, &y);
    if (p == 1) {
      printf("%d\n", bit.sum(y - 1) - bit.sum(x - 1) + 1);
    } else {
      set<int> s;
      if (r[a[x] - 1] > r[a[x]] && a[x] > 1) s.insert(a[x] - 1);
      if (r[a[x]] > r[a[x] + 1] && a[x] + 1 <= n) s.insert(a[x]);
      if (r[a[y] - 1] > r[a[y]] && a[y] > 1) s.insert(a[y] - 1);
      if (r[a[y]] > r[a[y] + 1] && a[y] + 1 <= n) s.insert(a[y]);
      for (auto &i : s) {
        bit.add(i, -1);
      }
      s.clear();
      swap(a[x], a[y]);
      swap(r[a[x]], r[a[y]]);
      if (r[a[x] - 1] > r[a[x]] && a[x] > 1) s.insert(a[x] - 1);
      if (r[a[x]] > r[a[x] + 1] && a[x] + 1 <= n) s.insert(a[x]);
      if (r[a[y] - 1] > r[a[y]] && a[y] > 1) s.insert(a[y] - 1);
      if (r[a[y]] > r[a[y] + 1] && a[y] + 1 <= n) s.insert(a[y]);
      for (auto &i : s) {
        bit.add(i, 1);
      }
      s.clear();
    }
  }
}

```