#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const unsigned gen_seed =
    std::chrono::system_clock::now().time_since_epoch().count();
std::mt19937_64 gen(gen_seed);
class stree {
 public:
  std::vector<int> t;
  int s;
  void build(std::vector<int>& a, int v, int tl, int tr) {
    if (v == 1) {
      t.resize(a.size() * 4);
      s = a.size();
    }
    if (tl == tr)
      t[v] = a[tl];
    else {
      int tm = (tl + tr) / 2;
      build(a, v * 2, tl, tm);
      build(a, v * 2 + 1, tm + 1, tr);
      t[v] = t[v * 2] + t[v * 2 + 1];
    }
  }
  void set(int v, int tl, int tr, int pos, int val) {
    if (pos == tl && tr == pos) {
      t[v] = val;
    } else {
      int tm = (tl + tr) / 2;
      if (pos <= tm)
        set(v * 2, tl, tm, pos, val);
      else
        set(v * 2 + 1, tm + 1, tr, pos, val);
      t[v] = t[v * 2] + t[v * 2 + 1];
    }
  }
  int get(int v, int tl, int tr, int l, int r) {
    if (l > r) return 0;
    if (tl == l && tr == r) return t[v];
    int tm = (tl + tr) / 2;
    return get(v * 2, tl, tm, l, min(r, tm)) +
           get(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
  }
};
stree st;
std::vector<int> bad;
std::vector<int> a, pos;
int n;
void upd(int i) {
  if (i == -1 || i == n - 1) return;
  int was = bad[i];
  int cur = 0;
  if (pos[i] > pos[i + 1]) cur = 1;
  if (was != cur) {
    bad[i] = cur;
    st.set(1, 0, n - 2, i, cur);
  }
}
int main() {
  scanf("%d", &n);
  a.resize(n);
  pos.resize(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
    pos[a[i]] = i;
  }
  bad = std::vector<int>(n - 1, 0);
  for (int i = 0; i < n - 1; i++)
    if (pos[i] > pos[i + 1]) bad[i] = 1;
  st.build(bad, 1, 0, n - 2);
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int t, l, r;
    scanf("%d %d %d", &t, &l, &r);
    l--;
    r--;
    if (t == 1)
      printf("%d\n", st.get(1, 0, n - 2, l, r - 1) + 1);
    else {
      swap(a[l], a[r]);
      pos[a[l]] = l;
      pos[a[r]] = r;
      upd(a[l] - 1);
      upd(a[l]);
      upd(a[r] - 1);
      upd(a[r]);
    }
  }
}

### 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;
const long long mod = 1000000007;
const unsigned gen_seed =
    std::chrono::system_clock::now().time_since_epoch().count();
std::mt19937_64 gen(gen_seed);
class stree {
 public:
  std::vector<int> t;
  int s;
  void build(std::vector<int>& a, int v, int tl, int tr) {
    if (v == 1) {
      t.resize(a.size() * 4);
      s = a.size();
    }
    if (tl == tr)
      t[v] = a[tl];
    else {
      int tm = (tl + tr) / 2;
      build(a, v * 2, tl, tm);
      build(a, v * 2 + 1, tm + 1, tr);
      t[v] = t[v * 2] + t[v * 2 + 1];
    }
  }
  void set(int v, int tl, int tr, int pos, int val) {
    if (pos == tl && tr == pos) {
      t[v] = val;
    } else {
      int tm = (tl + tr) / 2;
      if (pos <= tm)
        set(v * 2, tl, tm, pos, val);
      else
        set(v * 2 + 1, tm + 1, tr, pos, val);
      t[v] = t[v * 2] + t[v * 2 + 1];
    }
  }
  int get(int v, int tl, int tr, int l, int r) {
    if (l > r) return 0;
    if (tl == l && tr == r) return t[v];
    int tm = (tl + tr) / 2;
    return get(v * 2, tl, tm, l, min(r, tm)) +
           get(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
  }
};
stree st;
std::vector<int> bad;
std::vector<int> a, pos;
int n;
void upd(int i) {
  if (i == -1 || i == n - 1) return;
  int was = bad[i];
  int cur = 0;
  if (pos[i] > pos[i + 1]) cur = 1;
  if (was != cur) {
    bad[i] = cur;
    st.set(1, 0, n - 2, i, cur);
  }
}
int main() {
  scanf("%d", &n);
  a.resize(n);
  pos.resize(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
    pos[a[i]] = i;
  }
  bad = std::vector<int>(n - 1, 0);
  for (int i = 0; i < n - 1; i++)
    if (pos[i] > pos[i + 1]) bad[i] = 1;
  st.build(bad, 1, 0, n - 2);
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int t, l, r;
    scanf("%d %d %d", &t, &l, &r);
    l--;
    r--;
    if (t == 1)
      printf("%d\n", st.get(1, 0, n - 2, l, r - 1) + 1);
    else {
      swap(a[l], a[r]);
      pos[a[l]] = l;
      pos[a[r]] = r;
      upd(a[l] - 1);
      upd(a[l]);
      upd(a[r] - 1);
      upd(a[r]);
    }
  }
}

```