#include <bits/stdc++.h>
using namespace std;
int n, q;
int a[1000000], p[1000000], tr[1000000], b[1000000];
void upd(int pos, int del) {
  for (; pos < n; pos |= (pos + 1)) tr[pos] += del;
}
int sum(int r) {
  int s = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) s += tr[r];
  return s;
}
void UPD(int x) {
  if (x > 0 && p[x - 1] > p[x] && b[x - 1] == 0) {
    b[x - 1] = 1;
    upd(x - 1, 1);
  } else if (x > 0 && p[x - 1] < p[x] && b[x - 1] == 1) {
    b[x - 1] = 0;
    upd(x - 1, -1);
  }
  if (x < n - 1 && p[x] > p[x + 1] && b[x] == 0) {
    b[x] = 1;
    upd(x, 1);
  } else if (x < n - 1 && p[x] < p[x + 1] && b[x] == 1) {
    b[x] = 0;
    upd(x, -1);
  }
}
int main() {
  cin >> n;
  int x;
  for (int(i) = (0); i < (n); ++(i)) {
    scanf("%d", &x);
    --x;
    a[i] = x;
    p[x] = i;
  }
  for (int(i) = (0); i < (n - 1); ++(i))
    if (p[i] > p[i + 1]) {
      b[i] = 1;
      upd(i, 1);
    }
  cin >> q;
  for (int(i) = (0); i < (q); ++(i)) {
    int t, x, y;
    scanf("%d%d%d", &t, &x, &y);
    --x;
    --y;
    if (t == 1) {
      cout << sum(y - 1) - sum(x - 1) + 1 << '\n';
    } else {
      swap(a[x], a[y]);
      p[a[x]] = x;
      p[a[y]] = y;
      UPD(a[x]);
      UPD(a[y]);
    }
  }
  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 n, q;
int a[1000000], p[1000000], tr[1000000], b[1000000];
void upd(int pos, int del) {
  for (; pos < n; pos |= (pos + 1)) tr[pos] += del;
}
int sum(int r) {
  int s = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) s += tr[r];
  return s;
}
void UPD(int x) {
  if (x > 0 && p[x - 1] > p[x] && b[x - 1] == 0) {
    b[x - 1] = 1;
    upd(x - 1, 1);
  } else if (x > 0 && p[x - 1] < p[x] && b[x - 1] == 1) {
    b[x - 1] = 0;
    upd(x - 1, -1);
  }
  if (x < n - 1 && p[x] > p[x + 1] && b[x] == 0) {
    b[x] = 1;
    upd(x, 1);
  } else if (x < n - 1 && p[x] < p[x + 1] && b[x] == 1) {
    b[x] = 0;
    upd(x, -1);
  }
}
int main() {
  cin >> n;
  int x;
  for (int(i) = (0); i < (n); ++(i)) {
    scanf("%d", &x);
    --x;
    a[i] = x;
    p[x] = i;
  }
  for (int(i) = (0); i < (n - 1); ++(i))
    if (p[i] > p[i + 1]) {
      b[i] = 1;
      upd(i, 1);
    }
  cin >> q;
  for (int(i) = (0); i < (q); ++(i)) {
    int t, x, y;
    scanf("%d%d%d", &t, &x, &y);
    --x;
    --y;
    if (t == 1) {
      cout << sum(y - 1) - sum(x - 1) + 1 << '\n';
    } else {
      swap(a[x], a[y]);
      p[a[x]] = x;
      p[a[y]] = y;
      UPD(a[x]);
      UPD(a[y]);
    }
  }
  return 0;
}

```