#include <bits/stdc++.h>
using namespace std;
const int N = (int)3e5 + 10;
int n, op, x, y, q, c[N], a[N], f[N], m[N], p[N];
void add(int at, int by) {
  while (at < N) {
    f[at] += by;
    at |= (at + 1);
  }
}
int query(int at) {
  int ret = 0;
  while (at >= 0) {
    ret += f[at];
    at = (at & (at + 1)) - 1;
  }
  return ret;
}
void update(int id, int sgn, int at) {
  if (id < 0) {
    return;
  }
  if (m[id] == at) {
    return;
  }
  m[id] = at;
  if (p[id] > p[id + 1]) {
    c[id] = 1;
  } else {
    c[id] = 0;
  }
  if (sgn == -1) {
    add(id, -c[id]);
  } else {
    add(id, c[id]);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
    p[a[i]] = i;
  }
  memset(f, 0, sizeof f);
  for (int i = 0; i < n; i++) {
    if (p[i] > p[i + 1]) {
      c[i] = 1;
    } else {
      c[i] = 0;
    }
    add(i, c[i]);
  }
  int step = 0;
  memset(m, -1, sizeof m);
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d %d %d", &op, &x, &y);
    x--;
    y--;
    if (op == 1) {
      printf("%d\n", query(y - 1) - query(x - 1) + 1);
    } else {
      ++step;
      update(a[x] - 1, -1, step);
      update(a[x], -1, step);
      update(a[y] - 1, -1, step);
      update(a[y], -1, step);
      swap(a[x], a[y]);
      p[a[x]] = x;
      p[a[y]] = y;
      ++step;
      update(a[x] - 1, 1, step);
      update(a[x], 1, step);
      update(a[y] - 1, 1, step);
      update(a[y], 1, step);
    }
  }
  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;
const int N = (int)3e5 + 10;
int n, op, x, y, q, c[N], a[N], f[N], m[N], p[N];
void add(int at, int by) {
  while (at < N) {
    f[at] += by;
    at |= (at + 1);
  }
}
int query(int at) {
  int ret = 0;
  while (at >= 0) {
    ret += f[at];
    at = (at & (at + 1)) - 1;
  }
  return ret;
}
void update(int id, int sgn, int at) {
  if (id < 0) {
    return;
  }
  if (m[id] == at) {
    return;
  }
  m[id] = at;
  if (p[id] > p[id + 1]) {
    c[id] = 1;
  } else {
    c[id] = 0;
  }
  if (sgn == -1) {
    add(id, -c[id]);
  } else {
    add(id, c[id]);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
    p[a[i]] = i;
  }
  memset(f, 0, sizeof f);
  for (int i = 0; i < n; i++) {
    if (p[i] > p[i + 1]) {
      c[i] = 1;
    } else {
      c[i] = 0;
    }
    add(i, c[i]);
  }
  int step = 0;
  memset(m, -1, sizeof m);
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d %d %d", &op, &x, &y);
    x--;
    y--;
    if (op == 1) {
      printf("%d\n", query(y - 1) - query(x - 1) + 1);
    } else {
      ++step;
      update(a[x] - 1, -1, step);
      update(a[x], -1, step);
      update(a[y] - 1, -1, step);
      update(a[y], -1, step);
      swap(a[x], a[y]);
      p[a[x]] = x;
      p[a[y]] = y;
      ++step;
      update(a[x] - 1, 1, step);
      update(a[x], 1, step);
      update(a[y] - 1, 1, step);
      update(a[y], 1, step);
    }
  }
  return 0;
}

```