#include <bits/stdc++.h>
using namespace std;
void print(int *a, int n) {
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  printf("\n");
  return;
}
int p[400000];
int arr[400000];
int ans[1200000];
void build(int s, int e, int ind) {
  if (s == e) {
    ans[ind] = 0;
    return;
  }
  build(s, (s + e) / 2, ind * 2);
  build((s + e) / 2 + 1, e, ind * 2 + 1);
  ans[ind] =
      ans[ind * 2] + ans[ind * 2 + 1] + (p[(s + e) / 2 + 1] < p[(s + e) / 2]);
  return;
}
void update(int s, int e, int ind, int po) {
  if (s > po || e < po) return;
  if (s == e) return;
  update(s, (s + e) / 2, ind * 2, po);
  update((s + e) / 2 + 1, e, ind * 2 + 1, po);
  ans[ind] =
      ans[ind * 2] + ans[ind * 2 + 1] + (p[(s + e) / 2 + 1] < p[(s + e) / 2]);
  return;
}
int query(int s, int e, int ind, int x, int y) {
  if (e < x || s > y) return 0;
  if (s >= x && e <= y) return ans[ind];
  int sm = query(s, (s + e) / 2, ind * 2, x, y) +
           query((s + e) / 2 + 1, e, ind * 2 + 1, x, y);
  int m = (s + e) / 2;
  if (m >= x && m <= y && m + 1 >= x && m + 1 <= y) sm += (p[m + 1] < p[m]);
  return sm;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
    p[arr[i]] = i;
  }
  build(1, n, 1);
  int q;
  int t, x, y;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d %d %d", &t, &x, &y);
    if (t == 2) {
      swap(p[arr[x]], p[arr[y]]);
      swap(arr[x], arr[y]);
      update(1, n, 1, arr[x]);
      update(1, n, 1, arr[y]);
    } else {
      printf("%d\n", query(1, n, 1, x, y) + 1);
    }
  }
}

### Prompt

Create a solution in cpp for 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;
void print(int *a, int n) {
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  printf("\n");
  return;
}
int p[400000];
int arr[400000];
int ans[1200000];
void build(int s, int e, int ind) {
  if (s == e) {
    ans[ind] = 0;
    return;
  }
  build(s, (s + e) / 2, ind * 2);
  build((s + e) / 2 + 1, e, ind * 2 + 1);
  ans[ind] =
      ans[ind * 2] + ans[ind * 2 + 1] + (p[(s + e) / 2 + 1] < p[(s + e) / 2]);
  return;
}
void update(int s, int e, int ind, int po) {
  if (s > po || e < po) return;
  if (s == e) return;
  update(s, (s + e) / 2, ind * 2, po);
  update((s + e) / 2 + 1, e, ind * 2 + 1, po);
  ans[ind] =
      ans[ind * 2] + ans[ind * 2 + 1] + (p[(s + e) / 2 + 1] < p[(s + e) / 2]);
  return;
}
int query(int s, int e, int ind, int x, int y) {
  if (e < x || s > y) return 0;
  if (s >= x && e <= y) return ans[ind];
  int sm = query(s, (s + e) / 2, ind * 2, x, y) +
           query((s + e) / 2 + 1, e, ind * 2 + 1, x, y);
  int m = (s + e) / 2;
  if (m >= x && m <= y && m + 1 >= x && m + 1 <= y) sm += (p[m + 1] < p[m]);
  return sm;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
    p[arr[i]] = i;
  }
  build(1, n, 1);
  int q;
  int t, x, y;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d %d %d", &t, &x, &y);
    if (t == 2) {
      swap(p[arr[x]], p[arr[y]]);
      swap(arr[x], arr[y]);
      update(1, n, 1, arr[x]);
      update(1, n, 1, arr[y]);
    } else {
      printf("%d\n", query(1, n, 1, x, y) + 1);
    }
  }
}

```