#include <bits/stdc++.h>
using namespace std;
vector<int> fenv;
int sum(int r) {
  int ans = 0;
  while (r >= 0) {
    ans += fenv[r];
    r = (r & (r + 1)) - 1;
  }
  return ans;
}
int sum(int l, int r) { return sum(r) - sum(l - 1); }
void inc(int i, int val) {
  while (i < fenv.size()) {
    fenv[i] += val;
    i = i | (i + 1);
  }
}
int main() {
  int n;
  cin >> n;
  vector<int> perm(n);
  for (int i = 0; i < n; ++i) {
    cin >> perm[i];
    --perm[i];
  }
  vector<int> pos(n);
  for (int i = 0; i < n; ++i) pos[perm[i]] = i;
  fenv.assign(n - 1, 0);
  for (int i = 0; i < n - 1; ++i)
    if (pos[i] > pos[i + 1]) inc(i, 1);
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int r, x, y;
    cin >> r >> x >> y;
    --x, --y;
    if (r == 1)
      cout << sum(x, y - 1) + 1 << endl;
    else {
      if (perm[x] > 0 && x <= pos[perm[x] - 1] && y >= pos[perm[x] - 1] &&
          perm[x] - 1 != perm[y])
        inc(perm[x] - 1, -1);
      if (perm[x] < n - 1 && x <= pos[perm[x] + 1] && y >= pos[perm[x] + 1])
        inc(perm[x], 1);
      if (perm[y] > 0 && x <= pos[perm[y] - 1] && y >= pos[perm[y] - 1] &&
          perm[y] - 1 != perm[x])
        inc(perm[y] - 1, 1);
      if (perm[y] < n - 1 && x <= pos[perm[y] + 1] && y >= pos[perm[y] + 1])
        inc(perm[y], -1);
      pos[perm[x]] = y;
      pos[perm[y]] = x;
      swap(perm[x], perm[y]);
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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;
vector<int> fenv;
int sum(int r) {
  int ans = 0;
  while (r >= 0) {
    ans += fenv[r];
    r = (r & (r + 1)) - 1;
  }
  return ans;
}
int sum(int l, int r) { return sum(r) - sum(l - 1); }
void inc(int i, int val) {
  while (i < fenv.size()) {
    fenv[i] += val;
    i = i | (i + 1);
  }
}
int main() {
  int n;
  cin >> n;
  vector<int> perm(n);
  for (int i = 0; i < n; ++i) {
    cin >> perm[i];
    --perm[i];
  }
  vector<int> pos(n);
  for (int i = 0; i < n; ++i) pos[perm[i]] = i;
  fenv.assign(n - 1, 0);
  for (int i = 0; i < n - 1; ++i)
    if (pos[i] > pos[i + 1]) inc(i, 1);
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int r, x, y;
    cin >> r >> x >> y;
    --x, --y;
    if (r == 1)
      cout << sum(x, y - 1) + 1 << endl;
    else {
      if (perm[x] > 0 && x <= pos[perm[x] - 1] && y >= pos[perm[x] - 1] &&
          perm[x] - 1 != perm[y])
        inc(perm[x] - 1, -1);
      if (perm[x] < n - 1 && x <= pos[perm[x] + 1] && y >= pos[perm[x] + 1])
        inc(perm[x], 1);
      if (perm[y] > 0 && x <= pos[perm[y] - 1] && y >= pos[perm[y] - 1] &&
          perm[y] - 1 != perm[x])
        inc(perm[y] - 1, 1);
      if (perm[y] < n - 1 && x <= pos[perm[y] + 1] && y >= pos[perm[y] + 1])
        inc(perm[y], -1);
      pos[perm[x]] = y;
      pos[perm[y]] = x;
      swap(perm[x], perm[y]);
    }
  }
  return 0;
}

```