#include <bits/stdc++.h>
using namespace std;
int a[400000];
int pos[400000];
int d[1 << 19];
int lowbit(int x) { return (x & (x ^ (x - 1))); }
void add1(int x) {
  int i;
  for (i = x; i < 1 << 19; i += lowbit(i)) d[i]++;
}
void rem1(int x) {
  int i;
  for (i = x; i < 1 << 19; i += lowbit(i)) d[i]--;
}
int get1(int x) {
  int i, ret;
  ret = 0;
  for (i = x; i > 0; i -= lowbit(i)) ret += d[i];
  return ret;
}
int main() {
  int i, n, q, x, y, z, tmp;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  memset(d, 0, sizeof(d));
  for (i = 1; i < n; i++)
    if (pos[i] > pos[i + 1]) {
      add1(i);
    }
  scanf("%d", &q);
  for (i = 0; i < q; i++) {
    scanf("%d%d%d", &x, &y, &z);
    if (x == 1) {
      tmp = get1(z - 1) - get1(y - 1);
      printf("%d\n", tmp + 1);
    } else {
      y--;
      z--;
      if ((a[y] > 1) && (pos[a[y] - 1] > pos[a[y]])) rem1(a[y] - 1);
      if ((a[y] + 1 <= n) && (pos[a[y]] > pos[a[y] + 1])) rem1(a[y]);
      if ((a[z] > 1) && (pos[a[z] - 1] > pos[a[z]]) && (a[z] - 1 != a[y]))
        rem1(a[z] - 1);
      if ((a[z] + 1 <= n) && (pos[a[z]] > pos[a[z] + 1]) && (a[z] != a[y] - 1))
        rem1(a[z]);
      tmp = a[y];
      a[y] = a[z];
      a[z] = tmp;
      pos[a[y]] = y;
      pos[a[z]] = z;
      if ((a[y] > 1) && (pos[a[y] - 1] > pos[a[y]])) add1(a[y] - 1);
      if ((a[y] + 1 <= n) && (pos[a[y]] > pos[a[y] + 1])) add1(a[y]);
      if ((a[z] > 1) && (pos[a[z] - 1] > pos[a[z]]) && (a[z] - 1 != a[y]))
        add1(a[z] - 1);
      if ((a[z] + 1 <= n) && (pos[a[z]] > pos[a[z] + 1]) && (a[z] != a[y] - 1))
        add1(a[z]);
    }
  }
}

### 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[400000];
int pos[400000];
int d[1 << 19];
int lowbit(int x) { return (x & (x ^ (x - 1))); }
void add1(int x) {
  int i;
  for (i = x; i < 1 << 19; i += lowbit(i)) d[i]++;
}
void rem1(int x) {
  int i;
  for (i = x; i < 1 << 19; i += lowbit(i)) d[i]--;
}
int get1(int x) {
  int i, ret;
  ret = 0;
  for (i = x; i > 0; i -= lowbit(i)) ret += d[i];
  return ret;
}
int main() {
  int i, n, q, x, y, z, tmp;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  memset(d, 0, sizeof(d));
  for (i = 1; i < n; i++)
    if (pos[i] > pos[i + 1]) {
      add1(i);
    }
  scanf("%d", &q);
  for (i = 0; i < q; i++) {
    scanf("%d%d%d", &x, &y, &z);
    if (x == 1) {
      tmp = get1(z - 1) - get1(y - 1);
      printf("%d\n", tmp + 1);
    } else {
      y--;
      z--;
      if ((a[y] > 1) && (pos[a[y] - 1] > pos[a[y]])) rem1(a[y] - 1);
      if ((a[y] + 1 <= n) && (pos[a[y]] > pos[a[y] + 1])) rem1(a[y]);
      if ((a[z] > 1) && (pos[a[z] - 1] > pos[a[z]]) && (a[z] - 1 != a[y]))
        rem1(a[z] - 1);
      if ((a[z] + 1 <= n) && (pos[a[z]] > pos[a[z] + 1]) && (a[z] != a[y] - 1))
        rem1(a[z]);
      tmp = a[y];
      a[y] = a[z];
      a[z] = tmp;
      pos[a[y]] = y;
      pos[a[z]] = z;
      if ((a[y] > 1) && (pos[a[y] - 1] > pos[a[y]])) add1(a[y] - 1);
      if ((a[y] + 1 <= n) && (pos[a[y]] > pos[a[y] + 1])) add1(a[y]);
      if ((a[z] > 1) && (pos[a[z] - 1] > pos[a[z]]) && (a[z] - 1 != a[y]))
        add1(a[z] - 1);
      if ((a[z] + 1 <= n) && (pos[a[z]] > pos[a[z] + 1]) && (a[z] != a[y] - 1))
        add1(a[z]);
    }
  }
}

```