#include <bits/stdc++.h>
int a[300300];
int b[300300];
int t[300300];
int n, q;
void update(int u, const int &val) {
  while (1 <= u && u <= n) {
    t[u] += val;
    u = u + (u & (-u));
  }
}
int sum(int u) {
  int res = 0;
  while (1 <= u && u <= n) {
    res += t[u];
    u = u & (u - 1);
  }
  return (res);
}
void init(void) {
  int i, v;
  scanf("%d", &n);
  for (i = 1; i <= n; i = i + 1) {
    scanf("%d", &v);
    a[i] = v;
    b[v] = i;
  }
  memset(t, 0, sizeof t);
  for (i = 2; i <= n; i = i + 1)
    if (b[i] < b[i - 1]) update(i, 1);
}
void swap(int &a, int &b) {
  int sw;
  sw = a;
  a = b;
  b = sw;
}
void answer(void) {
  int i, t, x, y, r, t1, t2;
  bool px, py, px1, py1;
  scanf("%d", &q);
  for (i = 1; i <= q; i = i + 1) {
    scanf("%d", &t);
    scanf("%d", &x);
    scanf("%d", &y);
    if (t == 2) {
      t1 = a[x];
      t2 = a[y];
      px = (b[t1] < b[t1 - 1]);
      py = (b[t2] < b[t2 - 1]);
      px1 = (b[t1 + 1] < b[t1]);
      py1 = (b[t2 + 1] < b[t2]);
      b[t1] = y;
      b[t2] = x;
      swap(a[x], a[y]);
      if (b[t1] < b[t1 - 1] && !px) update(t1, 1);
      if (b[t1] > b[t1 - 1] && px) update(t1, -1);
      if (b[t1 + 1] < b[t1] && !px1) update(t1 + 1, 1);
      if (b[t1 + 1] > b[t1] && px1) update(t1 + 1, -1);
      if (t2 != t1 && t2 != t1 + 1) {
        if (b[t2] < b[t2 - 1] && !py) update(t2, 1);
        if (b[t2] > b[t2 - 1] && py) update(t2, -1);
      }
      if (t2 + 1 != t1 && t2 + 1 != t1 + 1) {
        if (b[t2 + 1] < b[t2] && !py1) update(t2 + 1, 1);
        if (b[t2 + 1] > b[t2] && py1) update(t2 + 1, -1);
      }
    } else {
      r = sum(y) - sum(x);
      printf("%d\n", r + 1);
    }
  }
}
int main(void) {
  init();
  answer();
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
int a[300300];
int b[300300];
int t[300300];
int n, q;
void update(int u, const int &val) {
  while (1 <= u && u <= n) {
    t[u] += val;
    u = u + (u & (-u));
  }
}
int sum(int u) {
  int res = 0;
  while (1 <= u && u <= n) {
    res += t[u];
    u = u & (u - 1);
  }
  return (res);
}
void init(void) {
  int i, v;
  scanf("%d", &n);
  for (i = 1; i <= n; i = i + 1) {
    scanf("%d", &v);
    a[i] = v;
    b[v] = i;
  }
  memset(t, 0, sizeof t);
  for (i = 2; i <= n; i = i + 1)
    if (b[i] < b[i - 1]) update(i, 1);
}
void swap(int &a, int &b) {
  int sw;
  sw = a;
  a = b;
  b = sw;
}
void answer(void) {
  int i, t, x, y, r, t1, t2;
  bool px, py, px1, py1;
  scanf("%d", &q);
  for (i = 1; i <= q; i = i + 1) {
    scanf("%d", &t);
    scanf("%d", &x);
    scanf("%d", &y);
    if (t == 2) {
      t1 = a[x];
      t2 = a[y];
      px = (b[t1] < b[t1 - 1]);
      py = (b[t2] < b[t2 - 1]);
      px1 = (b[t1 + 1] < b[t1]);
      py1 = (b[t2 + 1] < b[t2]);
      b[t1] = y;
      b[t2] = x;
      swap(a[x], a[y]);
      if (b[t1] < b[t1 - 1] && !px) update(t1, 1);
      if (b[t1] > b[t1 - 1] && px) update(t1, -1);
      if (b[t1 + 1] < b[t1] && !px1) update(t1 + 1, 1);
      if (b[t1 + 1] > b[t1] && px1) update(t1 + 1, -1);
      if (t2 != t1 && t2 != t1 + 1) {
        if (b[t2] < b[t2 - 1] && !py) update(t2, 1);
        if (b[t2] > b[t2 - 1] && py) update(t2, -1);
      }
      if (t2 + 1 != t1 && t2 + 1 != t1 + 1) {
        if (b[t2 + 1] < b[t2] && !py1) update(t2 + 1, 1);
        if (b[t2 + 1] > b[t2] && py1) update(t2 + 1, -1);
      }
    } else {
      r = sum(y) - sum(x);
      printf("%d\n", r + 1);
    }
  }
}
int main(void) {
  init();
  answer();
  return 0;
}

```