#include <bits/stdc++.h>
using namespace std;
int pos[301000], n, sum[301000], a[301000], mark[301000];
int Lowbit(int x) { return -x & x; }
void Add(int x, int num) {
  while (x <= n) {
    sum[x] += num;
    x = x + Lowbit(x);
  }
}
int Sum(int x) {
  int cal = 0;
  while (x > 0) {
    cal += sum[x];
    x = x - Lowbit(x);
  }
  return cal;
}
void update(int x, int y) {
  int delta = y - mark[x];
  mark[x] = y;
  Add(x, delta);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  memset(sum, 0, sizeof(sum));
  for (int i = 1; i <= n; i++) update(a[i], i < pos[a[i] - 1]);
  int m;
  cin >> m;
  while (m--) {
    int tp, i, j;
    cin >> tp >> i >> j;
    if (tp == 1)
      cout << Sum(j) - Sum(i) + 1 << endl;
    else {
      swap(a[i], a[j]);
      pos[a[i]] = i;
      pos[a[j]] = j;
      update(a[i], pos[a[i]] < pos[a[i] - 1]);
      if (a[i] < n) update(a[i] + 1, pos[a[i] + 1] < pos[a[i]]);
      update(a[j], pos[a[j]] < pos[a[j] - 1]);
      if (a[j] < n) update(a[j] + 1, pos[a[j] + 1] < pos[a[j]]);
    }
  }
}

### Prompt

Your challenge is to write 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 pos[301000], n, sum[301000], a[301000], mark[301000];
int Lowbit(int x) { return -x & x; }
void Add(int x, int num) {
  while (x <= n) {
    sum[x] += num;
    x = x + Lowbit(x);
  }
}
int Sum(int x) {
  int cal = 0;
  while (x > 0) {
    cal += sum[x];
    x = x - Lowbit(x);
  }
  return cal;
}
void update(int x, int y) {
  int delta = y - mark[x];
  mark[x] = y;
  Add(x, delta);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  memset(sum, 0, sizeof(sum));
  for (int i = 1; i <= n; i++) update(a[i], i < pos[a[i] - 1]);
  int m;
  cin >> m;
  while (m--) {
    int tp, i, j;
    cin >> tp >> i >> j;
    if (tp == 1)
      cout << Sum(j) - Sum(i) + 1 << endl;
    else {
      swap(a[i], a[j]);
      pos[a[i]] = i;
      pos[a[j]] = j;
      update(a[i], pos[a[i]] < pos[a[i] - 1]);
      if (a[i] < n) update(a[i] + 1, pos[a[i] + 1] < pos[a[i]]);
      update(a[j], pos[a[j]] < pos[a[j] - 1]);
      if (a[j] < n) update(a[j] + 1, pos[a[j] + 1] < pos[a[j]]);
    }
  }
}

```