#include <bits/stdc++.h>
using namespace std;
struct timer {
  time_t start;
  timer() { start = clock(); }
  ~timer() {
    cerr << 1. * (clock() - start) / CLOCKS_PER_SEC << " secs" << endl;
  }
};
int n, a[300010], pos[300010], q;
struct BIT {
  int N;
  long long *bit;
  void init(int n) {
    N = n;
    bit = new long long[N + 1];
    for (int i = 0; i < (N + 1); i++) bit[i] = 0;
  }
  long long sum(int i) {
    long long s = 0;
    while (i) {
      s += bit[i];
      i -= i & -i;
    }
    return s;
  }
  void add(int i, long long x) {
    i++;
    while (i <= N) {
      bit[i] += x;
      i += i & -i;
    }
  }
};
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  BIT B;
  B.init(n + 2);
  for (int i = 1; i < (n + 1); i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  pos[n + 1] = 1070000000LL;
  for (int i = 1; i < (n + 1); i++) {
    if (pos[i] < pos[i - 1]) {
      B.add(i, 1);
    }
  }
  cin >> q;
  for (int ii = 0; ii < (q); ii++) {
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) {
      cout << 1 + B.sum(y + 1) - B.sum(x + 1) << endl;
    } else {
      long long s;
      s = B.sum(a[x] + 1) - B.sum(a[x]);
      B.add(a[x], -s);
      s = B.sum(a[x] + 1 + 1) - B.sum(a[x] + 1);
      B.add(a[x] + 1, -s);
      s = B.sum(a[y] + 1) - B.sum(a[y]);
      B.add(a[y], -s);
      s = B.sum(a[y] + 1 + 1) - B.sum(a[y] + 1);
      B.add(a[y] + 1, -s);
      swap(pos[a[x]], pos[a[y]]);
      swap(a[x], a[y]);
      if (pos[a[x]] < pos[a[x] - 1]) {
        B.add(a[x], 1);
      }
      if (pos[a[x] + 1] < pos[a[x]]) {
        B.add(a[x] + 1, 1);
      }
      if (a[y] != a[x] + 1) {
        if (pos[a[y]] < pos[a[y] - 1]) {
          B.add(a[y], 1);
        }
      }
      if (a[y] + 1 != a[x]) {
        if (pos[a[y] + 1] < pos[a[y]]) {
          B.add(a[y] + 1, 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;
struct timer {
  time_t start;
  timer() { start = clock(); }
  ~timer() {
    cerr << 1. * (clock() - start) / CLOCKS_PER_SEC << " secs" << endl;
  }
};
int n, a[300010], pos[300010], q;
struct BIT {
  int N;
  long long *bit;
  void init(int n) {
    N = n;
    bit = new long long[N + 1];
    for (int i = 0; i < (N + 1); i++) bit[i] = 0;
  }
  long long sum(int i) {
    long long s = 0;
    while (i) {
      s += bit[i];
      i -= i & -i;
    }
    return s;
  }
  void add(int i, long long x) {
    i++;
    while (i <= N) {
      bit[i] += x;
      i += i & -i;
    }
  }
};
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  BIT B;
  B.init(n + 2);
  for (int i = 1; i < (n + 1); i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  pos[n + 1] = 1070000000LL;
  for (int i = 1; i < (n + 1); i++) {
    if (pos[i] < pos[i - 1]) {
      B.add(i, 1);
    }
  }
  cin >> q;
  for (int ii = 0; ii < (q); ii++) {
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) {
      cout << 1 + B.sum(y + 1) - B.sum(x + 1) << endl;
    } else {
      long long s;
      s = B.sum(a[x] + 1) - B.sum(a[x]);
      B.add(a[x], -s);
      s = B.sum(a[x] + 1 + 1) - B.sum(a[x] + 1);
      B.add(a[x] + 1, -s);
      s = B.sum(a[y] + 1) - B.sum(a[y]);
      B.add(a[y], -s);
      s = B.sum(a[y] + 1 + 1) - B.sum(a[y] + 1);
      B.add(a[y] + 1, -s);
      swap(pos[a[x]], pos[a[y]]);
      swap(a[x], a[y]);
      if (pos[a[x]] < pos[a[x] - 1]) {
        B.add(a[x], 1);
      }
      if (pos[a[x] + 1] < pos[a[x]]) {
        B.add(a[x] + 1, 1);
      }
      if (a[y] != a[x] + 1) {
        if (pos[a[y]] < pos[a[y] - 1]) {
          B.add(a[y], 1);
        }
      }
      if (a[y] + 1 != a[x]) {
        if (pos[a[y] + 1] < pos[a[y]]) {
          B.add(a[y] + 1, 1);
        }
      }
    }
  }
}

```