#include <bits/stdc++.h>
using namespace std;
int angka[300005], kiri, kanan, tanya, tipe, b1, b2;
int fenwick[300005], banyak;
int lokasi[300005];
void add(int x, int y) {
  while (x <= banyak) {
    fenwick[x] += y;
    x += (x & -x);
  }
}
int sum(int x) {
  int total = 0;
  while (x) {
    total += fenwick[x];
    x -= (x & -x);
  }
  return total;
}
int main() {
  scanf("%d", &banyak);
  for (int i = 1; i <= banyak; ++i) {
    scanf("%d", &angka[i]);
    lokasi[angka[i]] = i;
  }
  for (int i = 1; i < banyak; ++i) {
    if (lokasi[i] > lokasi[i + 1]) add(i, 1);
  }
  scanf("%d", &tanya);
  for (int l = 1; l <= tanya; ++l) {
    scanf("%d %d %d", &tipe, &kiri, &kanan);
    if (tipe == 1) {
      printf("%d\n", 1 + sum(kanan - 1) - sum(kiri - 1));
    } else {
      b1 = angka[kiri];
      b2 = angka[kanan];
      if (b1 > 1 && lokasi[b1 - 1] > lokasi[b1]) add(b1 - 1, -1);
      if (b1 < banyak && lokasi[b1] > lokasi[b1 + 1]) add(b1, -1);
      if (b2 > 1 && lokasi[b2 - 1] > lokasi[b2] && (b1 != b2 - 1))
        add(b2 - 1, -1);
      if (b2 < banyak && lokasi[b2] > lokasi[b2 + 1] && (b2 != b1 - 1))
        add(b2, -1);
      swap(angka[kiri], angka[kanan]);
      lokasi[b1] = kanan;
      lokasi[b2] = kiri;
      if (b1 > 1 && lokasi[b1 - 1] > lokasi[b1]) add(b1 - 1, 1);
      if (b1 < banyak && lokasi[b1] > lokasi[b1 + 1]) add(b1, 1);
      if (b2 > 1 && lokasi[b2 - 1] > lokasi[b2] && (b1 != b2 - 1))
        add(b2 - 1, 1);
      if (b2 < banyak && lokasi[b2] > lokasi[b2 + 1] && (b2 != b1 - 1))
        add(b2, 1);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to 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;
int angka[300005], kiri, kanan, tanya, tipe, b1, b2;
int fenwick[300005], banyak;
int lokasi[300005];
void add(int x, int y) {
  while (x <= banyak) {
    fenwick[x] += y;
    x += (x & -x);
  }
}
int sum(int x) {
  int total = 0;
  while (x) {
    total += fenwick[x];
    x -= (x & -x);
  }
  return total;
}
int main() {
  scanf("%d", &banyak);
  for (int i = 1; i <= banyak; ++i) {
    scanf("%d", &angka[i]);
    lokasi[angka[i]] = i;
  }
  for (int i = 1; i < banyak; ++i) {
    if (lokasi[i] > lokasi[i + 1]) add(i, 1);
  }
  scanf("%d", &tanya);
  for (int l = 1; l <= tanya; ++l) {
    scanf("%d %d %d", &tipe, &kiri, &kanan);
    if (tipe == 1) {
      printf("%d\n", 1 + sum(kanan - 1) - sum(kiri - 1));
    } else {
      b1 = angka[kiri];
      b2 = angka[kanan];
      if (b1 > 1 && lokasi[b1 - 1] > lokasi[b1]) add(b1 - 1, -1);
      if (b1 < banyak && lokasi[b1] > lokasi[b1 + 1]) add(b1, -1);
      if (b2 > 1 && lokasi[b2 - 1] > lokasi[b2] && (b1 != b2 - 1))
        add(b2 - 1, -1);
      if (b2 < banyak && lokasi[b2] > lokasi[b2 + 1] && (b2 != b1 - 1))
        add(b2, -1);
      swap(angka[kiri], angka[kanan]);
      lokasi[b1] = kanan;
      lokasi[b2] = kiri;
      if (b1 > 1 && lokasi[b1 - 1] > lokasi[b1]) add(b1 - 1, 1);
      if (b1 < banyak && lokasi[b1] > lokasi[b1 + 1]) add(b1, 1);
      if (b2 > 1 && lokasi[b2 - 1] > lokasi[b2] && (b1 != b2 - 1))
        add(b2 - 1, 1);
      if (b2 < banyak && lokasi[b2] > lokasi[b2 + 1] && (b2 != b1 - 1))
        add(b2, 1);
    }
  }
  return 0;
}

```