#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.);
const double EPS = 1e-8;
const int MAX_N = 1000000 + 10;
int N;
int b[MAX_N];
int a[MAX_N];
struct Tree {
  int l, r, val;
  Tree *ch[2];
  Tree() {}
  Tree(int _l, int _r, int *a) {
    l = _l;
    r = _r;
    if (l + 1 == r) {
      val = 1;
      return;
    }
    int mid = l + r >> 1;
    ch[0] = new Tree(l, mid, a);
    ch[1] = new Tree(mid, r, a);
    val = ch[0]->val + ch[1]->val;
    if (a[mid - 1] < a[mid]) --val;
  }
  int query(int _l, int _r) {
    if (_l <= l && _r >= r) return val;
    int mid = l + r >> 1;
    if (_r <= mid) return ch[0]->query(_l, _r);
    if (_l >= mid) return ch[1]->query(_l, _r);
    int ret = ch[0]->query(_l, _r) + ch[1]->query(_l, _r);
    if (a[mid - 1] < a[mid]) --ret;
    return ret;
  }
  void modify(int p) {
    if (p < l || p >= r) return;
    if (l + 1 == r) {
      val = 1;
      return;
    }
    ch[0]->modify(p);
    ch[1]->modify(p);
    val = ch[0]->val + ch[1]->val;
    int mid = l + r >> 1;
    if (a[mid - 1] < a[mid]) --val;
  }
};
void solve() {
  cin >> N;
  for (int i = 0; i < N; ++i) {
    scanf("%d", &b[i]);
    --b[i];
    a[b[i]] = i;
  }
  Tree *root = new Tree(0, N, a);
  int Q;
  cin >> Q;
  int opt, l, r;
  for (int i = 0; i < Q; ++i) {
    scanf("%d%d%d", &opt, &l, &r);
    if (opt == 1) {
      cout << root->query(l - 1, r) << endl;
    } else {
      --l, --r;
      swap(b[l], b[r]);
      a[b[l]] = l;
      a[b[r]] = r;
      root->modify(b[l]);
      root->modify(b[r]);
    }
  }
}
int main() {
  solve();
  return 0;
}

### Prompt

Please provide a CPP coded solution 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 double PI = acos(-1.);
const double EPS = 1e-8;
const int MAX_N = 1000000 + 10;
int N;
int b[MAX_N];
int a[MAX_N];
struct Tree {
  int l, r, val;
  Tree *ch[2];
  Tree() {}
  Tree(int _l, int _r, int *a) {
    l = _l;
    r = _r;
    if (l + 1 == r) {
      val = 1;
      return;
    }
    int mid = l + r >> 1;
    ch[0] = new Tree(l, mid, a);
    ch[1] = new Tree(mid, r, a);
    val = ch[0]->val + ch[1]->val;
    if (a[mid - 1] < a[mid]) --val;
  }
  int query(int _l, int _r) {
    if (_l <= l && _r >= r) return val;
    int mid = l + r >> 1;
    if (_r <= mid) return ch[0]->query(_l, _r);
    if (_l >= mid) return ch[1]->query(_l, _r);
    int ret = ch[0]->query(_l, _r) + ch[1]->query(_l, _r);
    if (a[mid - 1] < a[mid]) --ret;
    return ret;
  }
  void modify(int p) {
    if (p < l || p >= r) return;
    if (l + 1 == r) {
      val = 1;
      return;
    }
    ch[0]->modify(p);
    ch[1]->modify(p);
    val = ch[0]->val + ch[1]->val;
    int mid = l + r >> 1;
    if (a[mid - 1] < a[mid]) --val;
  }
};
void solve() {
  cin >> N;
  for (int i = 0; i < N; ++i) {
    scanf("%d", &b[i]);
    --b[i];
    a[b[i]] = i;
  }
  Tree *root = new Tree(0, N, a);
  int Q;
  cin >> Q;
  int opt, l, r;
  for (int i = 0; i < Q; ++i) {
    scanf("%d%d%d", &opt, &l, &r);
    if (opt == 1) {
      cout << root->query(l - 1, r) << endl;
    } else {
      --l, --r;
      swap(b[l], b[r]);
      a[b[l]] = l;
      a[b[r]] = r;
      root->modify(b[l]);
      root->modify(b[r]);
    }
  }
}
int main() {
  solve();
  return 0;
}

```