#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100000 * 3 + 10;
int f[MAXN], n, q, num[MAXN], pos[MAXN];
int CC;
int lowbit(int x) { return x & -x; }
void Add(int x, int d) {
  for (; x <= n; f[x] += d, x += lowbit(x))
    ;
}
int Getsum(int x) {
  int CC = 0;
  for (; x; CC += f[x], x -= lowbit(x))
    ;
  return CC;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &num[i]), pos[num[i]] = i;
  for (int i = 1; i < n; ++i)
    if (pos[i] > pos[i + 1]) Add(i, 1);
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    int ty, x, y;
    scanf("%d%d%d", &ty, &x, &y);
    if (ty == 1)
      cout << Getsum(y - 1) - Getsum(x - 1) + 1 << endl;
    else {
      int Max = max(num[x], num[y]), Min = min(num[x], num[y]);
      int xl = pos[Min - 1] > pos[Min], xr = pos[Min] > pos[Min + 1];
      int yl = pos[Max - 1] > pos[Max], yr = pos[Max] > pos[Max + 1];
      swap(num[x], num[y]);
      pos[num[x]] = x, pos[num[y]] = y;
      if (Min > 1) Add(Min - 1, (pos[Min - 1] > pos[Min]) - xl);
      if (Min < n) Add(Min, (pos[Min] > pos[Min + 1]) - xr);
      if (Max > 1 && Max - 1 != Min)
        Add(Max - 1, (pos[Max - 1] > pos[Max]) - yl);
      if (Max < n) Add(Max, (pos[Max] > pos[Max + 1]) - yr);
    }
  }
  fclose(stdin);
  fclose(stdout);
  return 0;
}

### Prompt

Please formulate 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 MAXN = 100000 * 3 + 10;
int f[MAXN], n, q, num[MAXN], pos[MAXN];
int CC;
int lowbit(int x) { return x & -x; }
void Add(int x, int d) {
  for (; x <= n; f[x] += d, x += lowbit(x))
    ;
}
int Getsum(int x) {
  int CC = 0;
  for (; x; CC += f[x], x -= lowbit(x))
    ;
  return CC;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &num[i]), pos[num[i]] = i;
  for (int i = 1; i < n; ++i)
    if (pos[i] > pos[i + 1]) Add(i, 1);
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    int ty, x, y;
    scanf("%d%d%d", &ty, &x, &y);
    if (ty == 1)
      cout << Getsum(y - 1) - Getsum(x - 1) + 1 << endl;
    else {
      int Max = max(num[x], num[y]), Min = min(num[x], num[y]);
      int xl = pos[Min - 1] > pos[Min], xr = pos[Min] > pos[Min + 1];
      int yl = pos[Max - 1] > pos[Max], yr = pos[Max] > pos[Max + 1];
      swap(num[x], num[y]);
      pos[num[x]] = x, pos[num[y]] = y;
      if (Min > 1) Add(Min - 1, (pos[Min - 1] > pos[Min]) - xl);
      if (Min < n) Add(Min, (pos[Min] > pos[Min + 1]) - xr);
      if (Max > 1 && Max - 1 != Min)
        Add(Max - 1, (pos[Max - 1] > pos[Max]) - yl);
      if (Max < n) Add(Max, (pos[Max] > pos[Max + 1]) - yr);
    }
  }
  fclose(stdin);
  fclose(stdout);
  return 0;
}

```