#include <bits/stdc++.h>
using namespace std;
char qq[400000];
pair<int, int> m[400000];
vector<int> co;
int kr[400000], su[400000];
int n, x, y, q, a;
void sum(int i, int k) {
  qq[i] = 1;
  while (i <= n) {
    su[i] += k;
    i = i | (i + 1);
  }
}
int ot(int i) {
  int o = 0;
  while (i >= 0) {
    o += su[i];
    i = (i & (i + 1)) - 1;
  }
  return o;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> kr[i];
    kr[i]--;
    m[i].first = kr[i];
    m[i].second = i;
  }
  sort(m, m + n);
  sum(0, 1);
  for (int i = 1; i < n; ++i) {
    if (m[i].second < m[i - 1].second) {
      sum(i, 1);
    }
  }
  cin >> q;
  for (int i = 0; i < q; ++i) {
    cin >> a >> x >> y;
    x--;
    y--;
    if (a == 1) {
      cout << ot(y) - ot(x) + 1 << endl;
    }
    if (a == 2) {
      int d, t;
      d = kr[x];
      t = kr[y];
      swap(kr[x], kr[y]);
      x = d;
      y = t;
      swap(m[x].second, m[y].second);
      for (int i = x; i < x + 2; ++i) {
        if (qq[i]) {
          sum(i, -1);
          qq[i] = 0;
        }
        if (m[i].second < m[i - 1].second) {
          sum(i, 1);
        }
      }
      for (int i = y; i < y + 2; ++i) {
        if (qq[i]) {
          sum(i, -1);
          qq[i] = 0;
        }
        if (m[i].second < m[i - 1].second) {
          sum(i, 1);
        }
      }
    }
  }
}

### Prompt

Create a solution in CPP for 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;
char qq[400000];
pair<int, int> m[400000];
vector<int> co;
int kr[400000], su[400000];
int n, x, y, q, a;
void sum(int i, int k) {
  qq[i] = 1;
  while (i <= n) {
    su[i] += k;
    i = i | (i + 1);
  }
}
int ot(int i) {
  int o = 0;
  while (i >= 0) {
    o += su[i];
    i = (i & (i + 1)) - 1;
  }
  return o;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> kr[i];
    kr[i]--;
    m[i].first = kr[i];
    m[i].second = i;
  }
  sort(m, m + n);
  sum(0, 1);
  for (int i = 1; i < n; ++i) {
    if (m[i].second < m[i - 1].second) {
      sum(i, 1);
    }
  }
  cin >> q;
  for (int i = 0; i < q; ++i) {
    cin >> a >> x >> y;
    x--;
    y--;
    if (a == 1) {
      cout << ot(y) - ot(x) + 1 << endl;
    }
    if (a == 2) {
      int d, t;
      d = kr[x];
      t = kr[y];
      swap(kr[x], kr[y]);
      x = d;
      y = t;
      swap(m[x].second, m[y].second);
      for (int i = x; i < x + 2; ++i) {
        if (qq[i]) {
          sum(i, -1);
          qq[i] = 0;
        }
        if (m[i].second < m[i - 1].second) {
          sum(i, 1);
        }
      }
      for (int i = y; i < y + 2; ++i) {
        if (qq[i]) {
          sum(i, -1);
          qq[i] = 0;
        }
        if (m[i].second < m[i - 1].second) {
          sum(i, 1);
        }
      }
    }
  }
}

```