#include <bits/stdc++.h>
using namespace std;
long n, ar[500000], ps[500000], q, tp, a, b, bd[500000], sb[500000];
long res, fb;
long gs(long v) {
  fb = v / 200;
  res = 0;
  for (int i = 0; i < fb; i++) res += sb[i];
  for (int i = fb * 200; i <= v; i++) res += bd[i];
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ar[i];
    ps[ar[i]] = i;
  }
  for (int i = 2; i <= n; i++)
    if (ps[i] < ps[i - 1]) {
      bd[i] = 1;
      sb[i / 200]++;
    }
  cin >> q;
  for (; q; --q) {
    cin >> tp;
    if (tp == 1) {
      cin >> a >> b;
      cout << gs(b) - gs(a) + 1 << endl;
    } else {
      cin >> a >> b;
      swap(ar[a], ar[b]);
      ps[ar[a]] = a;
      ps[ar[b]] = b;
      a = ar[a];
      b = ar[b];
      sb[a / 200] -= bd[a];
      sb[b / 200] -= bd[b];
      bd[a] = 0;
      bd[b] = 0;
      sb[(a + 1) / 200] -= bd[a + 1];
      sb[(b + 1) / 200] -= bd[b + 1];
      bd[a] = bd[b] = bd[a + 1] = bd[b + 1] = 0;
      if (a > 1 && ps[a] < ps[a - 1]) {
        bd[a] = 1;
        sb[a / 200]++;
      }
      if (a < n && ps[a] > ps[a + 1]) {
        bd[a + 1] = 1;
        sb[(a + 1) / 200]++;
      }
      if (b > 1 && ps[b] < ps[b - 1]) {
        sb[b / 200] += 1 - bd[b];
        bd[b] = 1;
      }
      if (b < n && ps[b] > ps[b + 1]) {
        sb[(b + 1) / 200] += 1 - bd[b + 1];
        bd[b + 1] = 1;
      }
    }
  }
  cin.get();
  cin.get();
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
long n, ar[500000], ps[500000], q, tp, a, b, bd[500000], sb[500000];
long res, fb;
long gs(long v) {
  fb = v / 200;
  res = 0;
  for (int i = 0; i < fb; i++) res += sb[i];
  for (int i = fb * 200; i <= v; i++) res += bd[i];
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ar[i];
    ps[ar[i]] = i;
  }
  for (int i = 2; i <= n; i++)
    if (ps[i] < ps[i - 1]) {
      bd[i] = 1;
      sb[i / 200]++;
    }
  cin >> q;
  for (; q; --q) {
    cin >> tp;
    if (tp == 1) {
      cin >> a >> b;
      cout << gs(b) - gs(a) + 1 << endl;
    } else {
      cin >> a >> b;
      swap(ar[a], ar[b]);
      ps[ar[a]] = a;
      ps[ar[b]] = b;
      a = ar[a];
      b = ar[b];
      sb[a / 200] -= bd[a];
      sb[b / 200] -= bd[b];
      bd[a] = 0;
      bd[b] = 0;
      sb[(a + 1) / 200] -= bd[a + 1];
      sb[(b + 1) / 200] -= bd[b + 1];
      bd[a] = bd[b] = bd[a + 1] = bd[b + 1] = 0;
      if (a > 1 && ps[a] < ps[a - 1]) {
        bd[a] = 1;
        sb[a / 200]++;
      }
      if (a < n && ps[a] > ps[a + 1]) {
        bd[a + 1] = 1;
        sb[(a + 1) / 200]++;
      }
      if (b > 1 && ps[b] < ps[b - 1]) {
        sb[b / 200] += 1 - bd[b];
        bd[b] = 1;
      }
      if (b < n && ps[b] > ps[b + 1]) {
        sb[(b + 1) / 200] += 1 - bd[b + 1];
        bd[b + 1] = 1;
      }
    }
  }
  cin.get();
  cin.get();
  return 0;
}

```