#include <bits/stdc++.h>
using namespace std;
struct Node {
  int a, b;
  Node *left, *right;
  int sum;
};
const int N = 300000 + 10;
Node pool[N << 1], *last;
int a[N], p[N];
void build(Node *&cur, int a, int b) {
  cur = last++;
  cur->a = a;
  cur->b = b;
  cur->sum = 0;
  if (a + 1 == b) return;
  build(cur->left, a, (a + b) / 2);
  build(cur->right, (a + b) / 2, b);
}
int A, B;
void insert(Node *cur) {
  if (cur->a + 1 == cur->b) {
    cur->sum += B;
    return;
  }
  if ((cur->a + cur->b) / 2 > A)
    insert(cur->left);
  else
    insert(cur->right);
  cur->sum = cur->left->sum + cur->right->sum;
}
int query(Node *cur) {
  if (cur->a >= A && cur->b <= B) return cur->sum;
  int ret = 0;
  if ((cur->a + cur->b) / 2 > A) ret += query(cur->left);
  if ((cur->a + cur->b) / 2 < B) ret += query(cur->right);
  return ret;
}
void print(Node *cur, int dep) {
  for (int i = 0; i < dep; ++i) putchar(' ');
  printf("[%d,%d):%d\n", cur->a, cur->b, cur->sum);
  if (cur->a + 1 == cur->b) return;
  print(cur->left, dep + 1);
  print(cur->right, dep + 1);
}
int main() {
  int n, m;
  scanf("%d", &n);
  a[0] = p[0] = 0;
  a[n + 1] = p[n + 1] = n + 1;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    p[a[i]] = i;
  }
  last = pool;
  Node *root;
  build(root, 0, n);
  for (int i = 1; i < n; ++i) {
    if (p[i] > p[i + 1]) {
      A = i;
      B = 1;
      insert(root);
    }
  }
  scanf("%d", &m);
  while (m--) {
    int x, y, z;
    scanf("%d%d%d", &z, &x, &y);
    if (z == 1) {
      A = x;
      B = y;
      printf("%d\n", query(root) + 1);
    } else {
      if (a[x] > a[y]) swap(x, y);
      if (a[x] + 1 == a[y]) {
        A = a[x] - 1;
        B = (p[a[x] - 1] > p[a[y]]) - (p[a[x] - 1] > p[a[x]]);
        if (B) insert(root);
        A = a[x];
        B = (p[a[y]] > p[a[x]]) - (p[a[x]] > p[a[y]]);
        if (B) insert(root);
        A = a[y];
        B = (p[a[x]] > p[a[y] + 1]) - (p[a[y]] > p[a[y] + 1]);
        if (B) insert(root);
      } else {
        A = a[x] - 1;
        B = (p[a[x] - 1] > p[a[y]]) - (p[a[x] - 1] > p[a[x]]);
        if (B) insert(root);
        A = a[x];
        B = (p[a[y]] > p[a[x] + 1]) - (p[a[x]] > p[a[x] + 1]);
        if (B) insert(root);
        A = a[y] - 1;
        B = (p[a[y] - 1] > p[a[x]]) - (p[a[y] - 1] > p[a[y]]);
        if (B) insert(root);
        A = a[y];
        B = (p[a[x]] > p[a[y] + 1]) - (p[a[y]] > p[a[y] + 1]);
        if (B) insert(root);
      }
      swap(a[x], a[y]);
      p[a[x]] = x;
      p[a[y]] = y;
    }
  }
  return 0;
}

### Prompt

Your task is to create 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;
struct Node {
  int a, b;
  Node *left, *right;
  int sum;
};
const int N = 300000 + 10;
Node pool[N << 1], *last;
int a[N], p[N];
void build(Node *&cur, int a, int b) {
  cur = last++;
  cur->a = a;
  cur->b = b;
  cur->sum = 0;
  if (a + 1 == b) return;
  build(cur->left, a, (a + b) / 2);
  build(cur->right, (a + b) / 2, b);
}
int A, B;
void insert(Node *cur) {
  if (cur->a + 1 == cur->b) {
    cur->sum += B;
    return;
  }
  if ((cur->a + cur->b) / 2 > A)
    insert(cur->left);
  else
    insert(cur->right);
  cur->sum = cur->left->sum + cur->right->sum;
}
int query(Node *cur) {
  if (cur->a >= A && cur->b <= B) return cur->sum;
  int ret = 0;
  if ((cur->a + cur->b) / 2 > A) ret += query(cur->left);
  if ((cur->a + cur->b) / 2 < B) ret += query(cur->right);
  return ret;
}
void print(Node *cur, int dep) {
  for (int i = 0; i < dep; ++i) putchar(' ');
  printf("[%d,%d):%d\n", cur->a, cur->b, cur->sum);
  if (cur->a + 1 == cur->b) return;
  print(cur->left, dep + 1);
  print(cur->right, dep + 1);
}
int main() {
  int n, m;
  scanf("%d", &n);
  a[0] = p[0] = 0;
  a[n + 1] = p[n + 1] = n + 1;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    p[a[i]] = i;
  }
  last = pool;
  Node *root;
  build(root, 0, n);
  for (int i = 1; i < n; ++i) {
    if (p[i] > p[i + 1]) {
      A = i;
      B = 1;
      insert(root);
    }
  }
  scanf("%d", &m);
  while (m--) {
    int x, y, z;
    scanf("%d%d%d", &z, &x, &y);
    if (z == 1) {
      A = x;
      B = y;
      printf("%d\n", query(root) + 1);
    } else {
      if (a[x] > a[y]) swap(x, y);
      if (a[x] + 1 == a[y]) {
        A = a[x] - 1;
        B = (p[a[x] - 1] > p[a[y]]) - (p[a[x] - 1] > p[a[x]]);
        if (B) insert(root);
        A = a[x];
        B = (p[a[y]] > p[a[x]]) - (p[a[x]] > p[a[y]]);
        if (B) insert(root);
        A = a[y];
        B = (p[a[x]] > p[a[y] + 1]) - (p[a[y]] > p[a[y] + 1]);
        if (B) insert(root);
      } else {
        A = a[x] - 1;
        B = (p[a[x] - 1] > p[a[y]]) - (p[a[x] - 1] > p[a[x]]);
        if (B) insert(root);
        A = a[x];
        B = (p[a[y]] > p[a[x] + 1]) - (p[a[x]] > p[a[x] + 1]);
        if (B) insert(root);
        A = a[y] - 1;
        B = (p[a[y] - 1] > p[a[x]]) - (p[a[y] - 1] > p[a[y]]);
        if (B) insert(root);
        A = a[y];
        B = (p[a[x]] > p[a[y] + 1]) - (p[a[y]] > p[a[y] + 1]);
        if (B) insert(root);
      }
      swap(a[x], a[y]);
      p[a[x]] = x;
      p[a[y]] = y;
    }
  }
  return 0;
}

```