#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T abs(T t) {
  return t < 0 ? -t : t;
}
const unsigned long long modn = 1000000007;
inline unsigned long long mod(unsigned long long x) { return x % modn; }
const int MAX = 300009;
int n;
int v[MAX], inv[MAX];
int bit[MAX];
int get(int i) {
  int sum = 0;
  i += 2;
  while (i) {
    sum += bit[i];
    i -= (i & -i);
  }
  return sum;
}
int get(int i, int j) { return get(j) - get(i - 1); }
void add(int i, int x) {
  i += 2;
  while (i < MAX) {
    bit[i] += x;
    i += (i & -i);
  }
}
void deb() {
  return;
  for (int i = 0; i < n; i++) printf("%d ", v[i] + 1);
  putchar('\n');
  for (int i = 0; i < n; i++) printf("%d ", get(i, i));
  putchar('\n');
  puts("----------");
}
int main() {
  int i, j, q, op, x, y;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &v[i]);
    v[i]--;
    inv[v[i]] = i;
  }
  for (i = 0; i < n - 1; i++)
    if (inv[i + 1] < inv[i]) add(i, 1);
  scanf("%d", &q);
  for (i = 0; i < q; i++) {
    scanf("%d %d %d", &op, &x, &y);
    x--;
    y--;
    if (op == 1)
      printf("%d\n", 1 + get(x, y - 1));
    else {
      if (abs(v[x] - v[y]) == 1) {
        if (v[y] == v[x] + 1) {
          deb();
          add(v[x], 1);
          if (v[y] < n - 1 && inv[v[y] + 1] > x && inv[v[y] + 1] < y)
            add(v[y], -1);
          if (v[x] > 0 && inv[v[x] - 1] > x && inv[v[x] - 1] < y)
            add(v[x] - 1, -1);
        } else {
          deb();
          add(v[y], -1);
          deb();
          if (v[x] < n - 1 && inv[v[x] + 1] > x && inv[v[x] + 1] < y)
            add(v[x], 1);
          if (v[y] > 0 && inv[v[y] - 1] > x && inv[v[y] - 1] < y)
            add(v[y] - 1, 1);
        }
      } else {
        deb();
        if (v[x] > 0 && inv[v[x] - 1] > x && inv[v[x] - 1] < y)
          add(v[x] - 1, -1);
        if (v[x] < n - 1 && inv[v[x] + 1] > x && inv[v[x] + 1] < y)
          add(v[x], 1);
        if (v[y] > 0 && inv[v[y] - 1] > x && inv[v[y] - 1] < y)
          add(v[y] - 1, 1);
        if (v[y] < n - 1 && inv[v[y] + 1] > x && inv[v[y] + 1] < y)
          add(v[y], -1);
      }
      inv[v[x]] = y;
      inv[v[y]] = x;
      swap(v[x], v[y]);
      deb();
    }
  }
}

### Prompt

Your task is to create a cpp solution to 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;
template <typename T>
inline T abs(T t) {
  return t < 0 ? -t : t;
}
const unsigned long long modn = 1000000007;
inline unsigned long long mod(unsigned long long x) { return x % modn; }
const int MAX = 300009;
int n;
int v[MAX], inv[MAX];
int bit[MAX];
int get(int i) {
  int sum = 0;
  i += 2;
  while (i) {
    sum += bit[i];
    i -= (i & -i);
  }
  return sum;
}
int get(int i, int j) { return get(j) - get(i - 1); }
void add(int i, int x) {
  i += 2;
  while (i < MAX) {
    bit[i] += x;
    i += (i & -i);
  }
}
void deb() {
  return;
  for (int i = 0; i < n; i++) printf("%d ", v[i] + 1);
  putchar('\n');
  for (int i = 0; i < n; i++) printf("%d ", get(i, i));
  putchar('\n');
  puts("----------");
}
int main() {
  int i, j, q, op, x, y;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &v[i]);
    v[i]--;
    inv[v[i]] = i;
  }
  for (i = 0; i < n - 1; i++)
    if (inv[i + 1] < inv[i]) add(i, 1);
  scanf("%d", &q);
  for (i = 0; i < q; i++) {
    scanf("%d %d %d", &op, &x, &y);
    x--;
    y--;
    if (op == 1)
      printf("%d\n", 1 + get(x, y - 1));
    else {
      if (abs(v[x] - v[y]) == 1) {
        if (v[y] == v[x] + 1) {
          deb();
          add(v[x], 1);
          if (v[y] < n - 1 && inv[v[y] + 1] > x && inv[v[y] + 1] < y)
            add(v[y], -1);
          if (v[x] > 0 && inv[v[x] - 1] > x && inv[v[x] - 1] < y)
            add(v[x] - 1, -1);
        } else {
          deb();
          add(v[y], -1);
          deb();
          if (v[x] < n - 1 && inv[v[x] + 1] > x && inv[v[x] + 1] < y)
            add(v[x], 1);
          if (v[y] > 0 && inv[v[y] - 1] > x && inv[v[y] - 1] < y)
            add(v[y] - 1, 1);
        }
      } else {
        deb();
        if (v[x] > 0 && inv[v[x] - 1] > x && inv[v[x] - 1] < y)
          add(v[x] - 1, -1);
        if (v[x] < n - 1 && inv[v[x] + 1] > x && inv[v[x] + 1] < y)
          add(v[x], 1);
        if (v[y] > 0 && inv[v[y] - 1] > x && inv[v[y] - 1] < y)
          add(v[y] - 1, 1);
        if (v[y] < n - 1 && inv[v[y] + 1] > x && inv[v[y] + 1] < y)
          add(v[y], -1);
      }
      inv[v[x]] = y;
      inv[v[y]] = x;
      swap(v[x], v[y]);
      deb();
    }
  }
}

```