#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:500000000")
using namespace std;
const int ppow = 1 << 19;
const int default_value = 0;
struct SegmentTree {
  int t[ppow * 2];
  SegmentTree() {
    for (int i = ppow * 2 - 1; i >= 0; i--) t[i] = default_value;
  }
  int F(int L, int R) {
    L += ppow;
    R += ppow;
    int res = default_value;
    while (L <= R) {
      if (L & 1) res = (res + t[L++]);
      if (~R & 1) res = (res + t[R--]);
      L /= 2;
      R /= 2;
    }
    return res;
  }
  void Add(int pos, int value) {
    pos += ppow;
    t[pos] += value;
    pos /= 2;
    while (pos > 0) {
      t[pos] = (t[pos * 2] + t[pos * 2 + 1]);
      pos /= 2;
    }
  }
  void Set(int pos, int value) {
    pos += ppow;
    t[pos] = value;
    pos /= 2;
    while (pos > 0) {
      t[pos] = (t[pos * 2] + t[pos * 2 + 1]);
      pos /= 2;
    }
  }
};
int A[1000009];
int Pos[1000009];
SegmentTree tree;
int main() {
  int n, i;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &A[i]);
    A[i]--;
    Pos[A[i]] = i;
  }
  for (i = 0; i < n - 1; i++)
    if (Pos[i] < Pos[i + 1]) tree.Set(i, 1);
  int q, p, x, y;
  cin >> q;
  while (q--) {
    scanf("%d %d %d", &p, &x, &y);
    x--, y--;
    if (p == 1) {
      printf("%d\n", y - x + 1 - tree.F(x, y - 1));
    } else {
      swap(A[x], A[y]);
      Pos[A[x]] = x;
      Pos[A[y]] = y;
      if (A[x] > 0) {
        int px = Pos[A[x] - 1];
        tree.Set(A[px], Pos[A[px]] < Pos[A[px] + 1]);
      }
      if (A[y] > 0) {
        int py = Pos[A[y] - 1];
        tree.Set(A[py], Pos[A[py]] < Pos[A[py] + 1]);
      }
      tree.Set(A[x], A[x] != n - 1 && Pos[A[x]] < Pos[A[x] + 1]);
      tree.Set(A[y], A[y] != n - 1 && Pos[A[y]] < Pos[A[y] + 1]);
    }
  }
}

### Prompt

Please formulate 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>
#pragma comment(linker, "/STACK:500000000")
using namespace std;
const int ppow = 1 << 19;
const int default_value = 0;
struct SegmentTree {
  int t[ppow * 2];
  SegmentTree() {
    for (int i = ppow * 2 - 1; i >= 0; i--) t[i] = default_value;
  }
  int F(int L, int R) {
    L += ppow;
    R += ppow;
    int res = default_value;
    while (L <= R) {
      if (L & 1) res = (res + t[L++]);
      if (~R & 1) res = (res + t[R--]);
      L /= 2;
      R /= 2;
    }
    return res;
  }
  void Add(int pos, int value) {
    pos += ppow;
    t[pos] += value;
    pos /= 2;
    while (pos > 0) {
      t[pos] = (t[pos * 2] + t[pos * 2 + 1]);
      pos /= 2;
    }
  }
  void Set(int pos, int value) {
    pos += ppow;
    t[pos] = value;
    pos /= 2;
    while (pos > 0) {
      t[pos] = (t[pos * 2] + t[pos * 2 + 1]);
      pos /= 2;
    }
  }
};
int A[1000009];
int Pos[1000009];
SegmentTree tree;
int main() {
  int n, i;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &A[i]);
    A[i]--;
    Pos[A[i]] = i;
  }
  for (i = 0; i < n - 1; i++)
    if (Pos[i] < Pos[i + 1]) tree.Set(i, 1);
  int q, p, x, y;
  cin >> q;
  while (q--) {
    scanf("%d %d %d", &p, &x, &y);
    x--, y--;
    if (p == 1) {
      printf("%d\n", y - x + 1 - tree.F(x, y - 1));
    } else {
      swap(A[x], A[y]);
      Pos[A[x]] = x;
      Pos[A[y]] = y;
      if (A[x] > 0) {
        int px = Pos[A[x] - 1];
        tree.Set(A[px], Pos[A[px]] < Pos[A[px] + 1]);
      }
      if (A[y] > 0) {
        int py = Pos[A[y] - 1];
        tree.Set(A[py], Pos[A[py]] < Pos[A[py] + 1]);
      }
      tree.Set(A[x], A[x] != n - 1 && Pos[A[x]] < Pos[A[x] + 1]);
      tree.Set(A[y], A[y] != n - 1 && Pos[A[y]] < Pos[A[y] + 1]);
    }
  }
}

```