#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3e5 + 6.66;
int seg[2 * MAXN];
int arr[MAXN];
int rev[MAXN];
int n;
void alter(int x, int val) {
  for (seg[x += n] = val; x > 1; x >>= 1) seg[x >> 1] = seg[x] + seg[x ^ 1];
}
int get(int l, int r) {
  int ans = 0;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) ans += seg[l++];
    if (r & 1) ans += seg[--r];
  }
  return ans;
}
void init(int n) {
  alter(1, 1);
  for (int i = 1; i < n; i++)
    if (rev[i] > rev[i + 1]) alter(i + 1, 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  for (int i = 1; i <= n; i++) rev[arr[i]] = i;
  init(n);
  int Q;
  cin >> Q;
  while (Q--) {
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) {
      cout << get(x + 1, y + 1) + 1 << endl;
    } else {
      if (arr[x] > arr[y]) swap(x, y);
      rev[arr[x]] = y;
      rev[arr[y]] = x;
      alter(arr[x], rev[arr[x]] < rev[arr[x] - 1]);
      alter(arr[y], rev[arr[y]] < rev[arr[y] - 1]);
      alter(arr[x] + 1, rev[arr[x] + 1] < rev[arr[x]]);
      alter(arr[y] + 1, rev[arr[y] + 1] < rev[arr[y]]);
      swap(arr[x], arr[y]);
    }
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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 + 6.66;
int seg[2 * MAXN];
int arr[MAXN];
int rev[MAXN];
int n;
void alter(int x, int val) {
  for (seg[x += n] = val; x > 1; x >>= 1) seg[x >> 1] = seg[x] + seg[x ^ 1];
}
int get(int l, int r) {
  int ans = 0;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) ans += seg[l++];
    if (r & 1) ans += seg[--r];
  }
  return ans;
}
void init(int n) {
  alter(1, 1);
  for (int i = 1; i < n; i++)
    if (rev[i] > rev[i + 1]) alter(i + 1, 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  for (int i = 1; i <= n; i++) rev[arr[i]] = i;
  init(n);
  int Q;
  cin >> Q;
  while (Q--) {
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) {
      cout << get(x + 1, y + 1) + 1 << endl;
    } else {
      if (arr[x] > arr[y]) swap(x, y);
      rev[arr[x]] = y;
      rev[arr[y]] = x;
      alter(arr[x], rev[arr[x]] < rev[arr[x] - 1]);
      alter(arr[y], rev[arr[y]] < rev[arr[y] - 1]);
      alter(arr[x] + 1, rev[arr[x] + 1] < rev[arr[x]]);
      alter(arr[y] + 1, rev[arr[y] + 1] < rev[arr[y]]);
      swap(arr[x], arr[y]);
    }
  }
  return 0;
}

```