#include <bits/stdc++.h>
using namespace std;
int a[500000], b[500000], dif[500000];
int n;
vector<int> T;
void update(int r, int x) {
  for (; r <= n; r |= (r + 1)) T[r] += x;
}
int get0(int l) {
  int res = 0;
  for (; l > 0; l = (l & (l + 1)) - 1) res += T[l];
  return res;
}
int get(int l, int r) { return get0(r) - get0(l - 1); }
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> b[i];
  for (int i = 1; i <= n; i++) a[b[i]] = i;
  for (int i = 1; i <= n; i++) dif[i] = (a[i] - a[i - 1] < 0);
  int q;
  cin >> q;
  T.assign(n + 1, 0);
  for (int i = 1; i <= n; i++) update(i, dif[i]);
  for (int i = 2; i <= n; i++) dif[i] = a[i] - a[i - 1];
  for (; q; --q) {
    int p, l, r;
    cin >> p >> l >> r;
    if (p == 1)
      cout << get(l + 1, r) + 1 << endl;
    else {
      int x = l, y = r;
      swap(b[x], b[y]);
      l = b[x], r = b[y];
      if (dif[l] < 0 && dif[l] - a[l] + a[r] >= 0)
        update(l, -1);
      else if (dif[l] >= 0 && dif[l] - a[l] + a[r] < 0)
        update(l, 1);
      if (dif[l + 1] < 0 && dif[l + 1] + a[l] - a[r] >= 0)
        update(l + 1, -1);
      else if (dif[l + 1] >= 0 && dif[l + 1] + a[l] - a[r] < 0)
        update(l + 1, 1);
      dif[l] = dif[l] - a[l] + a[r];
      dif[l + 1] = dif[l + 1] + a[l] - a[r];
      if (dif[r] < 0 && dif[r] - a[r] + a[l] >= 0)
        update(r, -1);
      else if (dif[r] >= 0 && dif[r] - a[r] + a[l] < 0)
        update(r, 1);
      if (r < n)
        if (dif[r + 1] < 0 && dif[r + 1] + a[r] - a[l] >= 0)
          update(r + 1, -1);
        else if (dif[r + 1] >= 0 && dif[r + 1] + a[r] - a[l] < 0)
          update(r + 1, 1);
      dif[r] = dif[r] - a[r] + a[l];
      if (r < n) dif[r + 1] = dif[r + 1] + a[r] - a[l];
      swap(a[l], a[r]);
    }
  }
  return 0;
}

### 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;
int a[500000], b[500000], dif[500000];
int n;
vector<int> T;
void update(int r, int x) {
  for (; r <= n; r |= (r + 1)) T[r] += x;
}
int get0(int l) {
  int res = 0;
  for (; l > 0; l = (l & (l + 1)) - 1) res += T[l];
  return res;
}
int get(int l, int r) { return get0(r) - get0(l - 1); }
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> b[i];
  for (int i = 1; i <= n; i++) a[b[i]] = i;
  for (int i = 1; i <= n; i++) dif[i] = (a[i] - a[i - 1] < 0);
  int q;
  cin >> q;
  T.assign(n + 1, 0);
  for (int i = 1; i <= n; i++) update(i, dif[i]);
  for (int i = 2; i <= n; i++) dif[i] = a[i] - a[i - 1];
  for (; q; --q) {
    int p, l, r;
    cin >> p >> l >> r;
    if (p == 1)
      cout << get(l + 1, r) + 1 << endl;
    else {
      int x = l, y = r;
      swap(b[x], b[y]);
      l = b[x], r = b[y];
      if (dif[l] < 0 && dif[l] - a[l] + a[r] >= 0)
        update(l, -1);
      else if (dif[l] >= 0 && dif[l] - a[l] + a[r] < 0)
        update(l, 1);
      if (dif[l + 1] < 0 && dif[l + 1] + a[l] - a[r] >= 0)
        update(l + 1, -1);
      else if (dif[l + 1] >= 0 && dif[l + 1] + a[l] - a[r] < 0)
        update(l + 1, 1);
      dif[l] = dif[l] - a[l] + a[r];
      dif[l + 1] = dif[l + 1] + a[l] - a[r];
      if (dif[r] < 0 && dif[r] - a[r] + a[l] >= 0)
        update(r, -1);
      else if (dif[r] >= 0 && dif[r] - a[r] + a[l] < 0)
        update(r, 1);
      if (r < n)
        if (dif[r + 1] < 0 && dif[r + 1] + a[r] - a[l] >= 0)
          update(r + 1, -1);
        else if (dif[r + 1] >= 0 && dif[r + 1] + a[r] - a[l] < 0)
          update(r + 1, 1);
      dif[r] = dif[r] - a[r] + a[l];
      if (r < n) dif[r + 1] = dif[r + 1] + a[r] - a[l];
      swap(a[l], a[r]);
    }
  }
  return 0;
}

```