#include <bits/stdc++.h>
using namespace std;
const int N = 300010;
int a[N], pos[N], c[N], is[N];
int n;
void add(int i, int k) {
  while (i <= n) {
    c[i] += k;
    i += i & -i;
  }
}
int query(int i) {
  int ans = 0;
  while (i) {
    ans += c[i];
    i -= i & -i;
  }
  return ans;
}
int main() {
  int i, j, m, op, L, R;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", a + i);
    pos[a[i]] = i;
  }
  is[1] = 0;
  for (i = 2; i <= n; i++) {
    if (pos[i] < pos[i - 1]) add(i, 1), is[i] = 1;
  }
  scanf("%d", &m);
  while (m--) {
    scanf("%d%d%d", &op, &L, &R);
    if (op == 1)
      printf("%d\n", query(R) - query(L) + 1);
    else {
      int num1 = a[L], num2 = a[R];
      add(num1, -is[num1]);
      is[num1] = 0;
      add(num2, -is[num2]);
      is[num2] = 0;
      swap(a[L], a[R]);
      swap(pos[num1], pos[num2]);
      add(num1 + 1, -is[num1 + 1]);
      is[num1 + 1] = 0;
      add(num2 + 1, -is[num2 + 1]);
      is[num2 + 1] = 0;
      if (num1 > 1 && pos[num1] < pos[num1 - 1]) add(num1, 1), is[num1] = 1;
      if (num2 > 1 && pos[num2] < pos[num2 - 1]) add(num2, 1), is[num2] = 1;
      if (num1 < n && pos[num1 + 1] < pos[num1] && is[num1 + 1] == 0)
        add(num1 + 1, 1), is[num1 + 1] = 1;
      if (num2 < n && pos[num2 + 1] < pos[num2])
        add(num2 + 1, 1), is[num2 + 1] = 1;
    }
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 N = 300010;
int a[N], pos[N], c[N], is[N];
int n;
void add(int i, int k) {
  while (i <= n) {
    c[i] += k;
    i += i & -i;
  }
}
int query(int i) {
  int ans = 0;
  while (i) {
    ans += c[i];
    i -= i & -i;
  }
  return ans;
}
int main() {
  int i, j, m, op, L, R;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", a + i);
    pos[a[i]] = i;
  }
  is[1] = 0;
  for (i = 2; i <= n; i++) {
    if (pos[i] < pos[i - 1]) add(i, 1), is[i] = 1;
  }
  scanf("%d", &m);
  while (m--) {
    scanf("%d%d%d", &op, &L, &R);
    if (op == 1)
      printf("%d\n", query(R) - query(L) + 1);
    else {
      int num1 = a[L], num2 = a[R];
      add(num1, -is[num1]);
      is[num1] = 0;
      add(num2, -is[num2]);
      is[num2] = 0;
      swap(a[L], a[R]);
      swap(pos[num1], pos[num2]);
      add(num1 + 1, -is[num1 + 1]);
      is[num1 + 1] = 0;
      add(num2 + 1, -is[num2 + 1]);
      is[num2 + 1] = 0;
      if (num1 > 1 && pos[num1] < pos[num1 - 1]) add(num1, 1), is[num1] = 1;
      if (num2 > 1 && pos[num2] < pos[num2 - 1]) add(num2, 1), is[num2] = 1;
      if (num1 < n && pos[num1 + 1] < pos[num1] && is[num1 + 1] == 0)
        add(num1 + 1, 1), is[num1 + 1] = 1;
      if (num2 < n && pos[num2 + 1] < pos[num2])
        add(num2 + 1, 1), is[num2 + 1] = 1;
    }
  }
  return 0;
}

```