#include <bits/stdc++.h>
using namespace std;
int f[555555], g[555555], pos[555555], i, j, ans, n, a[555555], q, t1, x, y;
void update(int x, int d) {
  while (x <= n) {
    f[x] += d;
    x = x | (x + 1);
  }
}
int count(int x) {
  int ans = 0;
  while (x > 0) {
    ans += f[x];
    x = x & (x + 1);
    x -= 1;
  }
  return ans;
}
int get(int l, int r) {
  if (l > r) return 0;
  return count(r) - count(l - 1);
}
void doit(int p) {
  if (p > 1) {
    if (g[p] != 0) update(p, -g[p]);
    g[p] = 0;
    if (pos[p] < pos[p - 1]) {
      g[p] = 1;
      update(p, g[p]);
    }
  }
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  for (i = 2; i <= n; ++i)
    if (pos[i] < pos[i - 1]) {
      g[i] = 1;
      update(i, g[i]);
    }
  scanf("%d", &q);
  while (q > 0) {
    q--;
    scanf("%d%d%d", &t1, &x, &y);
    if (t1 == 1) {
      ans = get(x + 1, y) + 1;
      cout << ans << endl;
    } else {
      swap(a[x], a[y]);
      pos[a[x]] = x;
      pos[a[y]] = y;
      for (i = -100; i <= 100; ++i) {
        if (x + i > 0 && x + i <= n) doit(x + i);
        if (y + i > 0 && y + i <= n) doit(y + i);
      }
    }
  }
}

### Prompt

Generate 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;
int f[555555], g[555555], pos[555555], i, j, ans, n, a[555555], q, t1, x, y;
void update(int x, int d) {
  while (x <= n) {
    f[x] += d;
    x = x | (x + 1);
  }
}
int count(int x) {
  int ans = 0;
  while (x > 0) {
    ans += f[x];
    x = x & (x + 1);
    x -= 1;
  }
  return ans;
}
int get(int l, int r) {
  if (l > r) return 0;
  return count(r) - count(l - 1);
}
void doit(int p) {
  if (p > 1) {
    if (g[p] != 0) update(p, -g[p]);
    g[p] = 0;
    if (pos[p] < pos[p - 1]) {
      g[p] = 1;
      update(p, g[p]);
    }
  }
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  for (i = 2; i <= n; ++i)
    if (pos[i] < pos[i - 1]) {
      g[i] = 1;
      update(i, g[i]);
    }
  scanf("%d", &q);
  while (q > 0) {
    q--;
    scanf("%d%d%d", &t1, &x, &y);
    if (t1 == 1) {
      ans = get(x + 1, y) + 1;
      cout << ans << endl;
    } else {
      swap(a[x], a[y]);
      pos[a[x]] = x;
      pos[a[y]] = y;
      for (i = -100; i <= 100; ++i) {
        if (x + i > 0 && x + i <= n) doit(x + i);
        if (y + i > 0 && y + i <= n) doit(y + i);
      }
    }
  }
}

```