#include <bits/stdc++.h>
using namespace std;
set<int> Set;
int n, w[300001], i, Q, r[300001], L[101], IT[1300001], SZ = 524288;
int sum(int a, int b) {
  a += SZ, b += SZ;
  int s = 0;
  while (a <= b) {
    if (a & 1) s += IT[a];
    if (!(b & 1)) s += IT[b];
    a = (a + 1) >> 1, b = (b - 1) >> 1;
  }
  return s;
}
void ins(int a, int b) {
  a += SZ;
  while (a) {
    IT[a] += b;
    a >>= 1;
  }
}
void Do(int a) {
  set<int>::iterator it = Set.lower_bound(a);
  int s, l = *it;
  if (it == Set.begin())
    s = 1;
  else {
    it--;
    s = *it;
    s++;
    it++;
  }
  if (s < a) {
    if (r[a - 1] > r[a]) {
      Set.insert(a - 1);
      ins(a, 1);
    }
  }
  if (a < l) {
    if (r[a] > r[a + 1]) {
      Set.insert(a);
      ins(a + 1, 1);
    }
  }
  if (s > 1 && r[s - 1] < r[s]) {
    it = Set.find(s - 1);
    Set.erase(it);
    ins(s, -1);
  }
  if (l < n && r[l] < r[l + 1]) {
    it = Set.find(l);
    Set.erase(l);
    ins(l + 1, -1);
  }
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
    r[w[i]] = i;
  }
  for (i = 1; i <= n; i++) {
    if (r[i] > r[i + 1]) {
      Set.insert(i);
      if (i != n) ins(i + 1, 1);
    }
  }
  scanf("%d", &Q);
  int a, b, c;
  while (Q--) {
    scanf("%d%d%d", &a, &b, &c);
    if (a == 1) {
      printf("%d\n", sum(b + 1, c) + 1);
    } else {
      swap(w[b], w[c]);
      swap(r[w[b]], r[w[c]]);
      Do(w[b]);
      Do(w[c]);
    }
  }
}

### 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;
set<int> Set;
int n, w[300001], i, Q, r[300001], L[101], IT[1300001], SZ = 524288;
int sum(int a, int b) {
  a += SZ, b += SZ;
  int s = 0;
  while (a <= b) {
    if (a & 1) s += IT[a];
    if (!(b & 1)) s += IT[b];
    a = (a + 1) >> 1, b = (b - 1) >> 1;
  }
  return s;
}
void ins(int a, int b) {
  a += SZ;
  while (a) {
    IT[a] += b;
    a >>= 1;
  }
}
void Do(int a) {
  set<int>::iterator it = Set.lower_bound(a);
  int s, l = *it;
  if (it == Set.begin())
    s = 1;
  else {
    it--;
    s = *it;
    s++;
    it++;
  }
  if (s < a) {
    if (r[a - 1] > r[a]) {
      Set.insert(a - 1);
      ins(a, 1);
    }
  }
  if (a < l) {
    if (r[a] > r[a + 1]) {
      Set.insert(a);
      ins(a + 1, 1);
    }
  }
  if (s > 1 && r[s - 1] < r[s]) {
    it = Set.find(s - 1);
    Set.erase(it);
    ins(s, -1);
  }
  if (l < n && r[l] < r[l + 1]) {
    it = Set.find(l);
    Set.erase(l);
    ins(l + 1, -1);
  }
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
    r[w[i]] = i;
  }
  for (i = 1; i <= n; i++) {
    if (r[i] > r[i + 1]) {
      Set.insert(i);
      if (i != n) ins(i + 1, 1);
    }
  }
  scanf("%d", &Q);
  int a, b, c;
  while (Q--) {
    scanf("%d%d%d", &a, &b, &c);
    if (a == 1) {
      printf("%d\n", sum(b + 1, c) + 1);
    } else {
      swap(w[b], w[c]);
      swap(r[w[b]], r[w[c]]);
      Do(w[b]);
      Do(w[c]);
    }
  }
}

```