#include <bits/stdc++.h>
using namespace std;
int n, i, q, p, x, y, to[310000], a[310000], b[310000], c[310000];
void read(int &x) {
  char k;
  for (k = getchar(); k <= 32; k = getchar())
    ;
  for (x = 0; '0' <= k; k = getchar()) x = x * 10 + k - '0';
}
void modify(int x, int y) {
  for (; x < n; x += x & -x) c[x] += y;
}
int get(int x) {
  int sum = 0;
  for (; x; x -= x & -x) sum += c[x];
  return sum;
}
int main() {
  read(n);
  for (i = 1; i <= n; i++) read(a[i]), to[a[i]] = i;
  for (i = 1; i < n; i++)
    if (to[i] > to[i + 1]) modify(i, 1), b[i] = 1;
  to[0] = 0;
  to[n + 1] = n + 1;
  read(q);
  while (q--) {
    read(p);
    read(x);
    read(y);
    if (p == 2) {
      if (b[a[x] - 1]) modify(a[x] - 1, -1), b[a[x] - 1] = 0;
      if (b[a[x]]) modify(a[x], -1), b[a[x]] = 0;
      if (b[a[y] - 1]) modify(a[y] - 1, -1), b[a[y] - 1] = 0;
      if (b[a[y]]) modify(a[y], -1), b[a[y]] = 0;
      swap(a[x], a[y]);
      swap(to[a[x]], to[a[y]]);
      if (!b[a[x] - 1] && to[a[x] - 1] > to[a[x]])
        modify(a[x] - 1, 1), b[a[x] - 1] = 1;
      if (!b[a[x]] && to[a[x]] > to[a[x] + 1]) modify(a[x], 1), b[a[x]] = 1;
      if (!b[a[y] - 1] && to[a[y] - 1] > to[a[y]])
        modify(a[y] - 1, 1), b[a[y] - 1] = 1;
      if (!b[a[y]] && to[a[y]] > to[a[y] + 1]) modify(a[y], 1), b[a[y]] = 1;
    } else
      printf("%d\n", get(y - 1) - get(x - 1) + 1);
  }
}

### 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;
int n, i, q, p, x, y, to[310000], a[310000], b[310000], c[310000];
void read(int &x) {
  char k;
  for (k = getchar(); k <= 32; k = getchar())
    ;
  for (x = 0; '0' <= k; k = getchar()) x = x * 10 + k - '0';
}
void modify(int x, int y) {
  for (; x < n; x += x & -x) c[x] += y;
}
int get(int x) {
  int sum = 0;
  for (; x; x -= x & -x) sum += c[x];
  return sum;
}
int main() {
  read(n);
  for (i = 1; i <= n; i++) read(a[i]), to[a[i]] = i;
  for (i = 1; i < n; i++)
    if (to[i] > to[i + 1]) modify(i, 1), b[i] = 1;
  to[0] = 0;
  to[n + 1] = n + 1;
  read(q);
  while (q--) {
    read(p);
    read(x);
    read(y);
    if (p == 2) {
      if (b[a[x] - 1]) modify(a[x] - 1, -1), b[a[x] - 1] = 0;
      if (b[a[x]]) modify(a[x], -1), b[a[x]] = 0;
      if (b[a[y] - 1]) modify(a[y] - 1, -1), b[a[y] - 1] = 0;
      if (b[a[y]]) modify(a[y], -1), b[a[y]] = 0;
      swap(a[x], a[y]);
      swap(to[a[x]], to[a[y]]);
      if (!b[a[x] - 1] && to[a[x] - 1] > to[a[x]])
        modify(a[x] - 1, 1), b[a[x] - 1] = 1;
      if (!b[a[x]] && to[a[x]] > to[a[x] + 1]) modify(a[x], 1), b[a[x]] = 1;
      if (!b[a[y] - 1] && to[a[y] - 1] > to[a[y]])
        modify(a[y] - 1, 1), b[a[y] - 1] = 1;
      if (!b[a[y]] && to[a[y]] > to[a[y] + 1]) modify(a[y], 1), b[a[y]] = 1;
    } else
      printf("%d\n", get(y - 1) - get(x - 1) + 1);
  }
}

```