#include <bits/stdc++.h>
using namespace std;
long long int num[300500];
long long int pos[300500];
long long int st[300500 * 4];
long long int n, N;
long long int recur(long long int v, long long int l, long long int r,
                    long long int left, long long int right) {
  if (right <= l || r <= left) return 0;
  if (left <= l && r <= right) return st[v];
  long long int mid = (l + r) >> 1;
  long long int ans = 0;
  ans += recur(v * 2 + 0, l, mid, left, right);
  ans += recur(v * 2 + 1, mid, r, left, right);
  return ans;
}
void change(long long int pos, long long int value) {
  pos += N;
  st[pos] = value;
  while (1 < pos) {
    pos /= 2;
    st[pos] = st[pos * 2 + 0] + st[pos * 2 + 1];
  }
}
int main() {
  cin >> n;
  for (N = 1; N <= n; N *= 2)
    ;
  for (long long int i = 0; i < ((long long int)(n + 1)); ++i)
    if (i) {
      cin >> num[i];
      pos[num[i]] = i;
    }
  for (long long int i = 0; i < ((long long int)(n)); ++i)
    if (i) {
      if (pos[i] < pos[i + 1])
        change(i, 0);
      else
        change(i, 1);
    }
  long long int q;
  cin >> q;
  for (long long int i = 0; i < ((long long int)(q)); ++i) {
    long long int type, x, y;
    cin >> type >> x >> y;
    if (type == 1) {
      cout << 1 + recur(1, 0, N, x, y) << endl;
    } else {
      long long int nx = num[x];
      long long int ny = num[y];
      swap(num[x], num[y]);
      swap(pos[nx], pos[ny]);
      for (long long int step = 0; step < ((long long int)(2)); ++step) {
        long long int t = step == 0 ? nx : ny;
        if (t != 1) change(t - 1, pos[t - 1] < pos[t] ? 0 : 1);
        if (t != n) change(t, pos[t] < pos[t + 1] ? 0 : 1);
      }
    }
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long int num[300500];
long long int pos[300500];
long long int st[300500 * 4];
long long int n, N;
long long int recur(long long int v, long long int l, long long int r,
                    long long int left, long long int right) {
  if (right <= l || r <= left) return 0;
  if (left <= l && r <= right) return st[v];
  long long int mid = (l + r) >> 1;
  long long int ans = 0;
  ans += recur(v * 2 + 0, l, mid, left, right);
  ans += recur(v * 2 + 1, mid, r, left, right);
  return ans;
}
void change(long long int pos, long long int value) {
  pos += N;
  st[pos] = value;
  while (1 < pos) {
    pos /= 2;
    st[pos] = st[pos * 2 + 0] + st[pos * 2 + 1];
  }
}
int main() {
  cin >> n;
  for (N = 1; N <= n; N *= 2)
    ;
  for (long long int i = 0; i < ((long long int)(n + 1)); ++i)
    if (i) {
      cin >> num[i];
      pos[num[i]] = i;
    }
  for (long long int i = 0; i < ((long long int)(n)); ++i)
    if (i) {
      if (pos[i] < pos[i + 1])
        change(i, 0);
      else
        change(i, 1);
    }
  long long int q;
  cin >> q;
  for (long long int i = 0; i < ((long long int)(q)); ++i) {
    long long int type, x, y;
    cin >> type >> x >> y;
    if (type == 1) {
      cout << 1 + recur(1, 0, N, x, y) << endl;
    } else {
      long long int nx = num[x];
      long long int ny = num[y];
      swap(num[x], num[y]);
      swap(pos[nx], pos[ny]);
      for (long long int step = 0; step < ((long long int)(2)); ++step) {
        long long int t = step == 0 ? nx : ny;
        if (t != 1) change(t - 1, pos[t - 1] < pos[t] ? 0 : 1);
        if (t != n) change(t, pos[t] < pos[t + 1] ? 0 : 1);
      }
    }
  }
}

```