#include <bits/stdc++.h>
using namespace std;
int ta[300003], a[300003], pos[300003];
int n, m;
inline void edit(int x, int v) {
  while (x <= n) {
    ta[x] += v;
    x += (x & (-x));
  }
}
inline int sum(int x) {
  int ret = 0;
  while (x) {
    ret += ta[x];
    x -= (x & (-x));
  }
  return ret;
}
void dbg() {
  printf("man:");
  for (int i = 1; i <= n; ++i) printf(" %d", a[i]);
  puts("");
  printf("ta1-mesum:");
  for (int i = 1; i <= n; ++i) printf(" %d", sum(i));
  puts("");
}
int main() {
  int q, l, r, op;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  for (int i = 1; i <= n; ++i) {
    if (a[i] - 1 != 0 && (pos[a[i]] < pos[a[i] - 1])) edit(a[i], 1);
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d", &op, &l, &r);
    if (l > r) swap(l, r);
    if (op == 1)
      printf("%d\n", sum(r) - sum(l) + 1);
    else {
      if (l == r) continue;
      if (a[l] - 1 >= 1 && pos[a[l]] < pos[a[l] - 1]) edit(a[l], -1);
      if (a[l] + 1 <= n && pos[a[l]] > pos[a[l] + 1]) edit(a[l] + 1, -1);
      if (a[r] - 1 >= 1 && pos[a[r]] < pos[a[r] - 1] && (a[r] - 1 != a[l]))
        edit(a[r], -1);
      if (a[r] + 1 <= n && pos[a[r]] > pos[a[r] + 1] && (a[r] + 1 != a[l]))
        edit(a[r] + 1, -1);
      swap(pos[a[l]], pos[a[r]]);
      if (a[l] - 1 >= 1 && pos[a[l]] < pos[a[l] - 1]) edit(a[l], 1);
      if (a[l] + 1 <= n && pos[a[l]] > pos[a[l] + 1]) edit(a[l] + 1, 1);
      if (a[r] - 1 >= 1 && pos[a[r]] < pos[a[r] - 1] && (a[r] - 1 != a[l]))
        edit(a[r], 1);
      if (a[r] + 1 <= n && pos[a[r]] > pos[a[r] + 1] && (a[r] + 1 != a[l]))
        edit(a[r] + 1, 1);
      swap(a[l], a[r]);
    }
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
int ta[300003], a[300003], pos[300003];
int n, m;
inline void edit(int x, int v) {
  while (x <= n) {
    ta[x] += v;
    x += (x & (-x));
  }
}
inline int sum(int x) {
  int ret = 0;
  while (x) {
    ret += ta[x];
    x -= (x & (-x));
  }
  return ret;
}
void dbg() {
  printf("man:");
  for (int i = 1; i <= n; ++i) printf(" %d", a[i]);
  puts("");
  printf("ta1-mesum:");
  for (int i = 1; i <= n; ++i) printf(" %d", sum(i));
  puts("");
}
int main() {
  int q, l, r, op;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  for (int i = 1; i <= n; ++i) {
    if (a[i] - 1 != 0 && (pos[a[i]] < pos[a[i] - 1])) edit(a[i], 1);
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d", &op, &l, &r);
    if (l > r) swap(l, r);
    if (op == 1)
      printf("%d\n", sum(r) - sum(l) + 1);
    else {
      if (l == r) continue;
      if (a[l] - 1 >= 1 && pos[a[l]] < pos[a[l] - 1]) edit(a[l], -1);
      if (a[l] + 1 <= n && pos[a[l]] > pos[a[l] + 1]) edit(a[l] + 1, -1);
      if (a[r] - 1 >= 1 && pos[a[r]] < pos[a[r] - 1] && (a[r] - 1 != a[l]))
        edit(a[r], -1);
      if (a[r] + 1 <= n && pos[a[r]] > pos[a[r] + 1] && (a[r] + 1 != a[l]))
        edit(a[r] + 1, -1);
      swap(pos[a[l]], pos[a[r]]);
      if (a[l] - 1 >= 1 && pos[a[l]] < pos[a[l] - 1]) edit(a[l], 1);
      if (a[l] + 1 <= n && pos[a[l]] > pos[a[l] + 1]) edit(a[l] + 1, 1);
      if (a[r] - 1 >= 1 && pos[a[r]] < pos[a[r] - 1] && (a[r] - 1 != a[l]))
        edit(a[r], 1);
      if (a[r] + 1 <= n && pos[a[r]] > pos[a[r] + 1] && (a[r] + 1 != a[l]))
        edit(a[r] + 1, 1);
      swap(a[l], a[r]);
    }
  }
  return 0;
}

```