#include <bits/stdc++.h>
using namespace std;
class ST {
  const int n;
  vector<int> v;
  void set(const int p, const int sl, const int sr, const int idx,
           const int val) {
    if (sl == sr) {
      v[p] = val;
      return;
    }
    const int sm = (sl + sr) >> 1;
    if (idx <= sm)
      set(p * 2 + 1, sl, sm, idx, val);
    else
      set(p * 2 + 2, sm + 1, sr, idx, val);
    v[p] = v[p * 2 + 1] + v[p * 2 + 2];
  }
  int sum(const int p, const int sl, const int sr, const int l, const int r) {
    if (sl == l && sr == r) return v[p];
    const int sm = (sl + sr) >> 1;
    int res = 0;
    if (l <= sm) res += sum(p * 2 + 1, sl, sm, l, min(sm, r));
    if (sm < r) res += sum(p * 2 + 2, sm + 1, sr, max(l, sm + 1), r);
    return res;
  }

 public:
  ST(const int n) : n(n), v(n * 4, 0) {}
  void set(const int idx, const int val) { set(0, 0, n - 1, idx, val); }
  int sum(const int l, const int r) { return sum(0, 0, n - 1, l, r); }
};
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> vv(n);
  vector<int> v(n);
  for (int i = 0; i < n; ++i) {
    cin >> vv[i];
    v[--vv[i]] = i;
  }
  ST st(n);
  for (int i = 1; i < n; ++i) {
    if (v[i] < v[i - 1]) st.set(i - 1, 1);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int p, x, y;
    cin >> p >> x >> y;
    --x;
    --y;
    if (p == 1) {
      cout << st.sum(x, y) - st.sum(y, y) + 1 << '\n';
    } else {
      swap(vv[x], vv[y]);
      x = vv[x];
      y = vv[y];
      if (x != 0 && v[x] < v[x - 1]) st.set(x - 1, 0);
      if (y != 0 && v[y] < v[y - 1]) st.set(y - 1, 0);
      if (x != n - 1 && v[x + 1] < v[x]) st.set(x, 0);
      if (y != n - 1 && v[y + 1] < v[y]) st.set(y, 0);
      swap(v[x], v[y]);
      if (x != 0 && v[x] < v[x - 1]) st.set(x - 1, 1);
      if (y != 0 && v[y] < v[y - 1]) st.set(y - 1, 1);
      if (x != n - 1 && v[x + 1] < v[x]) st.set(x, 1);
      if (y != n - 1 && v[y + 1] < v[y]) st.set(y, 1);
    }
  }
}

### Prompt

Your task is to create 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;
class ST {
  const int n;
  vector<int> v;
  void set(const int p, const int sl, const int sr, const int idx,
           const int val) {
    if (sl == sr) {
      v[p] = val;
      return;
    }
    const int sm = (sl + sr) >> 1;
    if (idx <= sm)
      set(p * 2 + 1, sl, sm, idx, val);
    else
      set(p * 2 + 2, sm + 1, sr, idx, val);
    v[p] = v[p * 2 + 1] + v[p * 2 + 2];
  }
  int sum(const int p, const int sl, const int sr, const int l, const int r) {
    if (sl == l && sr == r) return v[p];
    const int sm = (sl + sr) >> 1;
    int res = 0;
    if (l <= sm) res += sum(p * 2 + 1, sl, sm, l, min(sm, r));
    if (sm < r) res += sum(p * 2 + 2, sm + 1, sr, max(l, sm + 1), r);
    return res;
  }

 public:
  ST(const int n) : n(n), v(n * 4, 0) {}
  void set(const int idx, const int val) { set(0, 0, n - 1, idx, val); }
  int sum(const int l, const int r) { return sum(0, 0, n - 1, l, r); }
};
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> vv(n);
  vector<int> v(n);
  for (int i = 0; i < n; ++i) {
    cin >> vv[i];
    v[--vv[i]] = i;
  }
  ST st(n);
  for (int i = 1; i < n; ++i) {
    if (v[i] < v[i - 1]) st.set(i - 1, 1);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int p, x, y;
    cin >> p >> x >> y;
    --x;
    --y;
    if (p == 1) {
      cout << st.sum(x, y) - st.sum(y, y) + 1 << '\n';
    } else {
      swap(vv[x], vv[y]);
      x = vv[x];
      y = vv[y];
      if (x != 0 && v[x] < v[x - 1]) st.set(x - 1, 0);
      if (y != 0 && v[y] < v[y - 1]) st.set(y - 1, 0);
      if (x != n - 1 && v[x + 1] < v[x]) st.set(x, 0);
      if (y != n - 1 && v[y + 1] < v[y]) st.set(y, 0);
      swap(v[x], v[y]);
      if (x != 0 && v[x] < v[x - 1]) st.set(x - 1, 1);
      if (y != 0 && v[y] < v[y - 1]) st.set(y - 1, 1);
      if (x != n - 1 && v[x + 1] < v[x]) st.set(x, 1);
      if (y != n - 1 && v[y + 1] < v[y]) st.set(y, 1);
    }
  }
}

```