#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
class SBT {
  struct Node {
    int value, size;
    Node *c[2];
    Node(int v, int s, Node *p) : value(v), size(s) { c[0] = c[1] = p; }
    void rz() { size = c[0]->size + c[1]->size + 1; }
  } * root, *null;
  void rotate(Node *&t, bool d) {
    Node *p = t->c[d];
    t->c[d] = p->c[!d];
    p->c[!d] = t;
    t->rz();
    p->rz();
    t = p;
  }
  void maintain(Node *&t, bool d) {
    if (t == null) return;
    Node *&p = t->c[d];
    if (p->c[d]->size > t->c[!d]->size) {
      rotate(t, d);
    } else if (p->c[!d]->size > t->c[!d]->size) {
      rotate(p, !d);
      rotate(t, d);
    } else {
      return;
    }
    maintain(t->c[0], 0);
    maintain(t->c[1], 1);
    maintain(t, 0);
    maintain(t, 1);
  }
  void ins(Node *&t, int x) {
    if (t == null) {
      t = new Node(x, 1, null);
      return;
    }
    if (x == t->value) return;
    bool d = x > t->value;
    ins(t->c[d], x);
    maintain(t, d);
    t->rz();
  }
  void rmv(Node *&t, int x) {
    if (t == null) return;
    if (t->value == x) {
      if (t->c[1] == null) {
        delete t;
        t = t->c[0];
        return;
      }
      if (t->c[0] == null) {
        delete t;
        t = t->c[1];
        return;
      }
      Node *p = t->c[1];
      while (p->c[0] != null) p = p->c[0];
      t->value = p->value;
      rmv(t->c[1], p->value);
      maintain(t, 0);
    } else {
      bool d = x > t->value;
      rmv(t->c[d], x);
      maintain(t, !d);
    }
    t->rz();
  }
  int sel(Node *t, int k) {
    int r = t->c[0]->size;
    if (k == r) return t->value;
    if (k < r) return sel(t->c[0], k);
    return sel(t->c[1], k - r - 1);
  }
  int rnk(Node *t, int x) {
    if (t == null) return 0;
    int r = t->c[0]->size;
    if (x == t->value) return r;
    if (x < t->value) return rnk(t->c[0], x);
    return r + 1 + rnk(t->c[1], x);
  }
  void clr(Node *t) {
    if (t == null) return;
    clr(t->c[0]);
    clr(t->c[1]);
    delete t;
  }

 public:
  SBT() {
    null = new Node(0, 0, 0);
    null->c[0] = null->c[1] = null;
    root = null;
  }
  ~SBT() {
    clr(root);
    delete null;
  }
  void insert(int x) {
    if (count(x)) return;
    ins(root, x);
  }
  void remove(int x) {
    if (!count(x)) return;
    rmv(root, x);
  }
  int select(int k) {
    if (k > root->size) return inf;
    return sel(root, k - 1);
  }
  int rank(int x) { return rnk(root, x); }
  bool count(int x) {
    int p = rnk(root, x);
    return sel(root, p) == x;
  }
  int size() { return root->size; }
  void clear() {
    clr(root);
    root = null;
  }
} tree;
const int maxn = 300000 + 5;
int n, q;
int a[maxn];
int idx[maxn];
int main() {
  scanf("%d", &n);
  memset(idx, 0, sizeof(idx));
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    idx[a[i]] = i;
    if (!idx[a[i] - 1]) tree.insert(a[i]);
  }
  scanf("%d", &q);
  while (q--) {
    int p, x, y;
    scanf("%d%d%d", &p, &x, &y);
    if (p == 1) {
      int l = tree.rank(x), r = tree.rank(y);
      l -= !tree.count(x);
      r -= !tree.count(y);
      printf("%d\n", r - l + 1);
    } else {
      int xx = a[x], yy = a[y];
      int xl = idx[xx - 1], xr = idx[xx + 1];
      int yl = idx[yy - 1], yr = idx[yy + 1];
      swap(a[x], a[y]);
      swap(idx[xx], idx[yy]);
      if (x < xl && idx[xx] > idx[xx - 1]) tree.remove(xx);
      if (x < xr && idx[xx] > idx[xx + 1]) tree.insert(xx + 1);
      if (y > yl && idx[yy] < idx[yy - 1]) tree.insert(yy);
      if (y > yr && idx[yy] < idx[yy + 1]) tree.remove(yy + 1);
    }
  }
  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 inf = 0x3f3f3f3f;
class SBT {
  struct Node {
    int value, size;
    Node *c[2];
    Node(int v, int s, Node *p) : value(v), size(s) { c[0] = c[1] = p; }
    void rz() { size = c[0]->size + c[1]->size + 1; }
  } * root, *null;
  void rotate(Node *&t, bool d) {
    Node *p = t->c[d];
    t->c[d] = p->c[!d];
    p->c[!d] = t;
    t->rz();
    p->rz();
    t = p;
  }
  void maintain(Node *&t, bool d) {
    if (t == null) return;
    Node *&p = t->c[d];
    if (p->c[d]->size > t->c[!d]->size) {
      rotate(t, d);
    } else if (p->c[!d]->size > t->c[!d]->size) {
      rotate(p, !d);
      rotate(t, d);
    } else {
      return;
    }
    maintain(t->c[0], 0);
    maintain(t->c[1], 1);
    maintain(t, 0);
    maintain(t, 1);
  }
  void ins(Node *&t, int x) {
    if (t == null) {
      t = new Node(x, 1, null);
      return;
    }
    if (x == t->value) return;
    bool d = x > t->value;
    ins(t->c[d], x);
    maintain(t, d);
    t->rz();
  }
  void rmv(Node *&t, int x) {
    if (t == null) return;
    if (t->value == x) {
      if (t->c[1] == null) {
        delete t;
        t = t->c[0];
        return;
      }
      if (t->c[0] == null) {
        delete t;
        t = t->c[1];
        return;
      }
      Node *p = t->c[1];
      while (p->c[0] != null) p = p->c[0];
      t->value = p->value;
      rmv(t->c[1], p->value);
      maintain(t, 0);
    } else {
      bool d = x > t->value;
      rmv(t->c[d], x);
      maintain(t, !d);
    }
    t->rz();
  }
  int sel(Node *t, int k) {
    int r = t->c[0]->size;
    if (k == r) return t->value;
    if (k < r) return sel(t->c[0], k);
    return sel(t->c[1], k - r - 1);
  }
  int rnk(Node *t, int x) {
    if (t == null) return 0;
    int r = t->c[0]->size;
    if (x == t->value) return r;
    if (x < t->value) return rnk(t->c[0], x);
    return r + 1 + rnk(t->c[1], x);
  }
  void clr(Node *t) {
    if (t == null) return;
    clr(t->c[0]);
    clr(t->c[1]);
    delete t;
  }

 public:
  SBT() {
    null = new Node(0, 0, 0);
    null->c[0] = null->c[1] = null;
    root = null;
  }
  ~SBT() {
    clr(root);
    delete null;
  }
  void insert(int x) {
    if (count(x)) return;
    ins(root, x);
  }
  void remove(int x) {
    if (!count(x)) return;
    rmv(root, x);
  }
  int select(int k) {
    if (k > root->size) return inf;
    return sel(root, k - 1);
  }
  int rank(int x) { return rnk(root, x); }
  bool count(int x) {
    int p = rnk(root, x);
    return sel(root, p) == x;
  }
  int size() { return root->size; }
  void clear() {
    clr(root);
    root = null;
  }
} tree;
const int maxn = 300000 + 5;
int n, q;
int a[maxn];
int idx[maxn];
int main() {
  scanf("%d", &n);
  memset(idx, 0, sizeof(idx));
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    idx[a[i]] = i;
    if (!idx[a[i] - 1]) tree.insert(a[i]);
  }
  scanf("%d", &q);
  while (q--) {
    int p, x, y;
    scanf("%d%d%d", &p, &x, &y);
    if (p == 1) {
      int l = tree.rank(x), r = tree.rank(y);
      l -= !tree.count(x);
      r -= !tree.count(y);
      printf("%d\n", r - l + 1);
    } else {
      int xx = a[x], yy = a[y];
      int xl = idx[xx - 1], xr = idx[xx + 1];
      int yl = idx[yy - 1], yr = idx[yy + 1];
      swap(a[x], a[y]);
      swap(idx[xx], idx[yy]);
      if (x < xl && idx[xx] > idx[xx - 1]) tree.remove(xx);
      if (x < xr && idx[xx] > idx[xx + 1]) tree.insert(xx + 1);
      if (y > yl && idx[yy] < idx[yy - 1]) tree.insert(yy);
      if (y > yr && idx[yy] < idx[yy + 1]) tree.remove(yy + 1);
    }
  }
  return 0;
}

```