#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000")
using namespace std;
const long double pi = acos(-1.0);
const int N = (int)1e6;
int sum[4 * N], m, pos[N], a[N], n, q, x, y, cmd;
int get(int t, int l, int r, int x, int y) {
  if ((r <= x) || (y <= l)) return 0;
  if ((x <= l) && (r <= y)) return sum[t];
  int m = (l + r) / 2;
  return get(t * 2 + 1, l, m, x, y) + get(t * 2 + 2, m, r, x, y);
}
void update(int t, int l, int r, int x, int value) {
  if ((x < l) || (x >= r)) return;
  if (l == r - 1) {
    assert(x == l);
    sum[t] = value;
    return;
  }
  int m = (l + r) / 2;
  update(t * 2 + 1, l, m, x, value);
  update(t * 2 + 2, m, r, x, value);
  sum[t] = sum[t * 2 + 1] + sum[t * 2 + 2];
}
void check(int x) {
  if (x > 0) update(0, 0, n, x - 1, (int)(pos[x - 1] > pos[x]));
  if (x + 1 < n) update(0, 0, n, x, (int)(pos[x] > pos[x + 1]));
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &x);
    x--;
    pos[x] = i, a[i] = x;
  }
  for (int i = 0; i < n - 1; ++i)
    update(0, 0, n, i, (int)(pos[i] > pos[i + 1]));
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    scanf("%d%d%d", &cmd, &x, &y);
    --x, --y;
    if (cmd == 1)
      printf("%d\n", 1 + get(0, 0, n, x, y));
    else {
      swap(a[x], a[y]);
      pos[a[x]] = x;
      pos[a[y]] = y;
      check(a[x]);
      check(a[y]);
    }
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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>
#pragma comment(linker, "/STACK:100000000")
using namespace std;
const long double pi = acos(-1.0);
const int N = (int)1e6;
int sum[4 * N], m, pos[N], a[N], n, q, x, y, cmd;
int get(int t, int l, int r, int x, int y) {
  if ((r <= x) || (y <= l)) return 0;
  if ((x <= l) && (r <= y)) return sum[t];
  int m = (l + r) / 2;
  return get(t * 2 + 1, l, m, x, y) + get(t * 2 + 2, m, r, x, y);
}
void update(int t, int l, int r, int x, int value) {
  if ((x < l) || (x >= r)) return;
  if (l == r - 1) {
    assert(x == l);
    sum[t] = value;
    return;
  }
  int m = (l + r) / 2;
  update(t * 2 + 1, l, m, x, value);
  update(t * 2 + 2, m, r, x, value);
  sum[t] = sum[t * 2 + 1] + sum[t * 2 + 2];
}
void check(int x) {
  if (x > 0) update(0, 0, n, x - 1, (int)(pos[x - 1] > pos[x]));
  if (x + 1 < n) update(0, 0, n, x, (int)(pos[x] > pos[x + 1]));
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &x);
    x--;
    pos[x] = i, a[i] = x;
  }
  for (int i = 0; i < n - 1; ++i)
    update(0, 0, n, i, (int)(pos[i] > pos[i + 1]));
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    scanf("%d%d%d", &cmd, &x, &y);
    --x, --y;
    if (cmd == 1)
      printf("%d\n", 1 + get(0, 0, n, x, y));
    else {
      swap(a[x], a[y]);
      pos[a[x]] = x;
      pos[a[y]] = y;
      check(a[x]);
      check(a[y]);
    }
  }
  return 0;
}

```