#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
const long long INF = 1e9 + 19;
int n, q;
int p[N];
int rev[N];
int tree[N];
void add(int pos, int val) {
  for (; pos <= n; pos |= (pos + 1)) tree[pos + 1] += val;
}
int get(int r) {
  int sum = 0;
  for (; r > 0; r &= (r - 1)) sum += tree[r];
  return sum;
}
int get(int l, int r) { return get(r) - get(l); }
void f(int val, int cof) {
  if (0 <= val && val + 1 < n && rev[val] > rev[val + 1]) add(val, cof);
}
void read() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    p[i]--;
  }
  for (int i = 0; i < n; i++) rev[p[i]] = i;
  for (int i = 0; i < n; i++) f(i, 1);
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int type, x, y;
    scanf("%d%d%d", &type, &x, &y);
    x--;
    y--;
    if (type == 1) {
      printf("%d\n", get(x, y) + 1);
    } else {
      assert(type == 2);
      vector<int> tmp = {p[x], p[x] - 1, p[y], p[y] - 1};
      sort((tmp).begin(), (tmp).end());
      tmp.resize(unique((tmp).begin(), (tmp).end()) - tmp.begin());
      for (auto w : tmp) f(w, -1);
      swap(p[x], p[y]);
      swap(rev[p[x]], rev[p[y]]);
      for (auto w : tmp) f(w, 1);
    }
  }
}
void solve() {}
void stress() {}
int main() {
  if (1) {
    read();
    solve();
  } else {
    stress();
  }
  return 0;
}

### 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 int N = 1e6 + 10;
const long long INF = 1e9 + 19;
int n, q;
int p[N];
int rev[N];
int tree[N];
void add(int pos, int val) {
  for (; pos <= n; pos |= (pos + 1)) tree[pos + 1] += val;
}
int get(int r) {
  int sum = 0;
  for (; r > 0; r &= (r - 1)) sum += tree[r];
  return sum;
}
int get(int l, int r) { return get(r) - get(l); }
void f(int val, int cof) {
  if (0 <= val && val + 1 < n && rev[val] > rev[val + 1]) add(val, cof);
}
void read() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    p[i]--;
  }
  for (int i = 0; i < n; i++) rev[p[i]] = i;
  for (int i = 0; i < n; i++) f(i, 1);
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int type, x, y;
    scanf("%d%d%d", &type, &x, &y);
    x--;
    y--;
    if (type == 1) {
      printf("%d\n", get(x, y) + 1);
    } else {
      assert(type == 2);
      vector<int> tmp = {p[x], p[x] - 1, p[y], p[y] - 1};
      sort((tmp).begin(), (tmp).end());
      tmp.resize(unique((tmp).begin(), (tmp).end()) - tmp.begin());
      for (auto w : tmp) f(w, -1);
      swap(p[x], p[y]);
      swap(rev[p[x]], rev[p[y]]);
      for (auto w : tmp) f(w, 1);
    }
  }
}
void solve() {}
void stress() {}
int main() {
  if (1) {
    read();
    solve();
  } else {
    stress();
  }
  return 0;
}

```