#include <bits/stdc++.h>
using namespace std;
const int inf = 3000009;
int pos[inf];
bool st[inf];
int tree[inf];
int num[inf];
void update(int a, int v) {
  if (!a) return;
  for (; a < inf; a += a & -a) tree[a] += v;
}
int query(int a) {
  int x = 0;
  for (; a; a -= a & -a) x += tree[a];
  return x;
}
int main() {
  int n, m, a, b, c;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &num[i]);
    pos[num[i]] = i;
  }
  pos[n + 1] = n + 3;
  for (int i = 1; i < n; ++i)
    if (pos[i] > pos[i + 1]) update(i, 1), st[i] = true;
  scanf("%d", &m);
  bool is;
  while (m--) {
    scanf("%d%d%d", &c, &a, &b);
    if (c == 1)
      printf("%d\n", query(b - 1) - query(a - 1) + 1);
    else {
      swap(num[a], num[b]);
      a = num[a], b = num[b];
      swap(pos[a], pos[b]);
      is = pos[a + 1] < pos[a];
      if (is != st[a]) update(a, is ? 1 : -1);
      st[a] = is;
      is = pos[b + 1] < pos[b];
      if (is != st[b]) update(b, is ? 1 : -1);
      st[b] = is;
      is = pos[b] < pos[b - 1];
      if (is != st[b - 1]) update(b - 1, is ? 1 : -1);
      st[b - 1] = is;
      is = pos[a] < pos[a - 1];
      if (is != st[a - 1]) update(a - 1, is ? 1 : -1);
      st[a - 1] = is;
    }
  }
}

### Prompt

In Cpp, your task is to solve 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 inf = 3000009;
int pos[inf];
bool st[inf];
int tree[inf];
int num[inf];
void update(int a, int v) {
  if (!a) return;
  for (; a < inf; a += a & -a) tree[a] += v;
}
int query(int a) {
  int x = 0;
  for (; a; a -= a & -a) x += tree[a];
  return x;
}
int main() {
  int n, m, a, b, c;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &num[i]);
    pos[num[i]] = i;
  }
  pos[n + 1] = n + 3;
  for (int i = 1; i < n; ++i)
    if (pos[i] > pos[i + 1]) update(i, 1), st[i] = true;
  scanf("%d", &m);
  bool is;
  while (m--) {
    scanf("%d%d%d", &c, &a, &b);
    if (c == 1)
      printf("%d\n", query(b - 1) - query(a - 1) + 1);
    else {
      swap(num[a], num[b]);
      a = num[a], b = num[b];
      swap(pos[a], pos[b]);
      is = pos[a + 1] < pos[a];
      if (is != st[a]) update(a, is ? 1 : -1);
      st[a] = is;
      is = pos[b + 1] < pos[b];
      if (is != st[b]) update(b, is ? 1 : -1);
      st[b] = is;
      is = pos[b] < pos[b - 1];
      if (is != st[b - 1]) update(b - 1, is ? 1 : -1);
      st[b - 1] = is;
      is = pos[a] < pos[a - 1];
      if (is != st[a - 1]) update(a - 1, is ? 1 : -1);
      st[a - 1] = is;
    }
  }
}

```