#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 5;
const int inf = 1e9;
const int mod = 1e9 + 7;
int p[maxn], ind[maxn];
bool mark[maxn];
int fen[maxn];
int n;
void add(int x, int val) {
  for (; x <= n; x += x & -x) {
    fen[x] += val;
  }
}
int get(int x) {
  int sum = 0;
  for (; x > 0; x -= x & -x) {
    sum += fen[x];
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int m;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    ind[p[i]] = i;
  }
  cin >> m;
  ind[n + 1] = inf;
  for (int i = 1; i <= n; i++) {
    if (ind[i] > ind[i + 1]) {
      add(i, 1);
      mark[i] = true;
    }
  }
  while (m--) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r;
      cin >> l >> r;
      cout << get(r) - get(l - 1) + (ind[r] < ind[r + 1]) << endl;
    }
    if (t == 2) {
      int a, b;
      cin >> a >> b;
      swap(p[a], p[b]);
      ind[p[a]] = a;
      ind[p[b]] = b;
      add(p[b], (b > ind[p[b] + 1]) - mark[p[b]]);
      mark[p[b]] = (b > ind[p[b] + 1]);
      add(p[a], (a > ind[p[a] + 1]) - mark[p[a]]);
      mark[p[a]] = (a > ind[p[a] + 1]);
      if (p[b] > 1) add(p[b] - 1, (ind[p[b] - 1] > b) - mark[p[b] - 1]);
      mark[p[b] - 1] = (ind[p[b] - 1] > b);
      if (p[a] > 1) add(p[a] - 1, (ind[p[a] - 1] > a) - mark[p[a] - 1]);
      mark[p[a] - 1] = (ind[p[a] - 1] > a);
    }
  }
}

### Prompt

Your task is to create 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 maxn = 3e5 + 5;
const int inf = 1e9;
const int mod = 1e9 + 7;
int p[maxn], ind[maxn];
bool mark[maxn];
int fen[maxn];
int n;
void add(int x, int val) {
  for (; x <= n; x += x & -x) {
    fen[x] += val;
  }
}
int get(int x) {
  int sum = 0;
  for (; x > 0; x -= x & -x) {
    sum += fen[x];
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int m;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    ind[p[i]] = i;
  }
  cin >> m;
  ind[n + 1] = inf;
  for (int i = 1; i <= n; i++) {
    if (ind[i] > ind[i + 1]) {
      add(i, 1);
      mark[i] = true;
    }
  }
  while (m--) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r;
      cin >> l >> r;
      cout << get(r) - get(l - 1) + (ind[r] < ind[r + 1]) << endl;
    }
    if (t == 2) {
      int a, b;
      cin >> a >> b;
      swap(p[a], p[b]);
      ind[p[a]] = a;
      ind[p[b]] = b;
      add(p[b], (b > ind[p[b] + 1]) - mark[p[b]]);
      mark[p[b]] = (b > ind[p[b] + 1]);
      add(p[a], (a > ind[p[a] + 1]) - mark[p[a]]);
      mark[p[a]] = (a > ind[p[a] + 1]);
      if (p[b] > 1) add(p[b] - 1, (ind[p[b] - 1] > b) - mark[p[b] - 1]);
      mark[p[b] - 1] = (ind[p[b] - 1] > b);
      if (p[a] > 1) add(p[a] - 1, (ind[p[a] - 1] > a) - mark[p[a] - 1]);
      mark[p[a] - 1] = (ind[p[a] - 1] > a);
    }
  }
}

```