#include <bits/stdc++.h>
using namespace std;
const int N = 300000, N_ = N - 1;
int aa[N], ii[N];
int st[N_ * 2], n_;
void swap_(int i, int j) {
  int a = aa[i], b = aa[j];
  aa[i] = b, aa[j] = a;
  ii[a] = j, ii[b] = i;
}
void pul(int i) { st[i] = st[i << 1] + st[i << 1 | 1]; }
void build(int n) {
  n_ = n - 1;
  for (int i = 0; i < n_; i++) st[n_ + i] = ii[i] > ii[i + 1];
  for (int i = n_ - 1; i; i--) pul(i);
}
void update(int i) {
  st[i += n_] ^= 1;
  while (i > 1) pul(i >>= 1);
}
int query(int l, int r) {
  int x = 0;
  for (l += n_, r += n_; l <= r; l >>= 1, r >>= 1) {
    if ((l & 1) == 1) x += st[l++];
    if ((r & 1) == 0) x += st[r--];
  }
  return x;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a, a--;
    aa[ii[a] = i] = a;
  }
  build(n);
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int a, b;
      cin >> a >> b, a--, b--;
      cout << query(a, b - 1) + 1 << '\n';
    } else {
      int i, j;
      cin >> i >> j, i--, j--;
      int a = aa[i], b = aa[j];
      if (a > b) {
        swap(a, b);
        swap(i, j);
      }
      if (a > 0 && (ii[a - 1] < i) != (ii[a - 1] < j)) update(a - 1);
      if (a + 1 == b)
        update(a);
      else {
        if (a + 1 < n && (i < ii[a + 1]) != (j < ii[a + 1])) update(a);
        if (b > 0 && (ii[b - 1] < j) != (ii[b - 1] < i))
          if (b - 1 != a) update(b - 1);
      }
      if (b + 1 < n && (j < ii[b + 1]) != (i < ii[b + 1])) update(b);
      swap_(i, j);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 300000, N_ = N - 1;
int aa[N], ii[N];
int st[N_ * 2], n_;
void swap_(int i, int j) {
  int a = aa[i], b = aa[j];
  aa[i] = b, aa[j] = a;
  ii[a] = j, ii[b] = i;
}
void pul(int i) { st[i] = st[i << 1] + st[i << 1 | 1]; }
void build(int n) {
  n_ = n - 1;
  for (int i = 0; i < n_; i++) st[n_ + i] = ii[i] > ii[i + 1];
  for (int i = n_ - 1; i; i--) pul(i);
}
void update(int i) {
  st[i += n_] ^= 1;
  while (i > 1) pul(i >>= 1);
}
int query(int l, int r) {
  int x = 0;
  for (l += n_, r += n_; l <= r; l >>= 1, r >>= 1) {
    if ((l & 1) == 1) x += st[l++];
    if ((r & 1) == 0) x += st[r--];
  }
  return x;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a, a--;
    aa[ii[a] = i] = a;
  }
  build(n);
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int a, b;
      cin >> a >> b, a--, b--;
      cout << query(a, b - 1) + 1 << '\n';
    } else {
      int i, j;
      cin >> i >> j, i--, j--;
      int a = aa[i], b = aa[j];
      if (a > b) {
        swap(a, b);
        swap(i, j);
      }
      if (a > 0 && (ii[a - 1] < i) != (ii[a - 1] < j)) update(a - 1);
      if (a + 1 == b)
        update(a);
      else {
        if (a + 1 < n && (i < ii[a + 1]) != (j < ii[a + 1])) update(a);
        if (b > 0 && (ii[b - 1] < j) != (ii[b - 1] < i))
          if (b - 1 != a) update(b - 1);
      }
      if (b + 1 < n && (j < ii[b + 1]) != (i < ii[b + 1])) update(b);
      swap_(i, j);
    }
  }
  return 0;
}

```