#include <bits/stdc++.h>
using namespace std;
const int MAX = 3e5 + 9;
long long a[MAX], n, q, p, x, y, f[MAX], pos[MAX], tmp[MAX];
void upd(int x, long long vv) {
  long long v = vv - tmp[x];
  tmp[x] = vv;
  for (; x <= n; x += x & -x) f[x] += v;
}
long long gt(long long x) {
  long long ret = 0;
  for (; x; x -= x & -x) ret += f[x];
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], pos[a[i]] = i;
  for (int i = 1; i <= n; i++)
    if (i < pos[a[i] - 1])
      upd(a[i], +1);
    else
      upd(a[i], 0);
  cin >> q;
  while (q--) {
    cin >> p >> x >> y;
    if (p == 1)
      cout << 1 + gt(y) - gt(x) << '\n';
    else {
      swap(a[x], a[y]), pos[a[x]] = x, pos[a[y]] = y;
      if (pos[a[x]] < pos[a[x] - 1])
        upd(a[x], +1);
      else
        upd(a[x], 0);
      if (a[x] != n)
        if (pos[a[x] + 1] < pos[a[x]])
          upd(a[x] + 1, +1);
        else
          upd(a[x] + 1, 0);
      if (pos[a[y]] < pos[a[y] - 1])
        upd(a[y], +1);
      else
        upd(a[y], 0);
      if (a[y] != n)
        if (pos[a[y] + 1] < pos[a[y]])
          upd(a[y] + 1, +1);
        else
          upd(a[y] + 1, 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 MAX = 3e5 + 9;
long long a[MAX], n, q, p, x, y, f[MAX], pos[MAX], tmp[MAX];
void upd(int x, long long vv) {
  long long v = vv - tmp[x];
  tmp[x] = vv;
  for (; x <= n; x += x & -x) f[x] += v;
}
long long gt(long long x) {
  long long ret = 0;
  for (; x; x -= x & -x) ret += f[x];
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], pos[a[i]] = i;
  for (int i = 1; i <= n; i++)
    if (i < pos[a[i] - 1])
      upd(a[i], +1);
    else
      upd(a[i], 0);
  cin >> q;
  while (q--) {
    cin >> p >> x >> y;
    if (p == 1)
      cout << 1 + gt(y) - gt(x) << '\n';
    else {
      swap(a[x], a[y]), pos[a[x]] = x, pos[a[y]] = y;
      if (pos[a[x]] < pos[a[x] - 1])
        upd(a[x], +1);
      else
        upd(a[x], 0);
      if (a[x] != n)
        if (pos[a[x] + 1] < pos[a[x]])
          upd(a[x] + 1, +1);
        else
          upd(a[x] + 1, 0);
      if (pos[a[y]] < pos[a[y] - 1])
        upd(a[y], +1);
      else
        upd(a[y], 0);
      if (a[y] != n)
        if (pos[a[y] + 1] < pos[a[y]])
          upd(a[y] + 1, +1);
        else
          upd(a[y] + 1, 0);
    }
  }
}

```