#include <bits/stdc++.h>
using namespace std;
int r[600500], s[600500], e[600500], x, y, q, u, v, n, m, w, t, o, l, k, p, j;
void upd(int x, int v) {
  for (x; x <= n; x = (x | (x - 1)) + 1) s[x] += v;
}
int sum(int x) {
  int r = 0;
  for (x; x; x &= x - 1) r += s[x];
  return r;
}
void calc(int &a, int &b, int &c, int &d) {
  a = ((x < n) && (r[x] > r[x + 1]));
  b = ((x > 1) && (r[x - 1] > r[x]));
  c = ((y < n) && (r[y] > r[y + 1]));
  d = ((y > 1) && (r[y - 1] > r[y]));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &e[i]);
    r[e[i]] = i;
  }
  for (int i = 1; i < n; i++) upd(i, r[i] > r[i + 1]);
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &q, &x, &y);
    if (q == 1)
      printf("%d\n", 1 + sum(y - 1) - sum(x - 1));
    else {
      swap(e[x], e[y]);
      x = e[x];
      y = e[y];
      if (x > y) swap(x, y);
      calc(q, w, j, t);
      swap(r[x], r[y]);
      calc(o, p, l, k);
      upd(x, o - q);
      upd(y, l - j);
      if (x > 1) upd(x - 1, p - w);
      if ((y > 1) && (y - 1 != x)) upd(y - 1, k - t);
    }
  }
}

### 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>
using namespace std;
int r[600500], s[600500], e[600500], x, y, q, u, v, n, m, w, t, o, l, k, p, j;
void upd(int x, int v) {
  for (x; x <= n; x = (x | (x - 1)) + 1) s[x] += v;
}
int sum(int x) {
  int r = 0;
  for (x; x; x &= x - 1) r += s[x];
  return r;
}
void calc(int &a, int &b, int &c, int &d) {
  a = ((x < n) && (r[x] > r[x + 1]));
  b = ((x > 1) && (r[x - 1] > r[x]));
  c = ((y < n) && (r[y] > r[y + 1]));
  d = ((y > 1) && (r[y - 1] > r[y]));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &e[i]);
    r[e[i]] = i;
  }
  for (int i = 1; i < n; i++) upd(i, r[i] > r[i + 1]);
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &q, &x, &y);
    if (q == 1)
      printf("%d\n", 1 + sum(y - 1) - sum(x - 1));
    else {
      swap(e[x], e[y]);
      x = e[x];
      y = e[y];
      if (x > y) swap(x, y);
      calc(q, w, j, t);
      swap(r[x], r[y]);
      calc(o, p, l, k);
      upd(x, o - q);
      upd(y, l - j);
      if (x > 1) upd(x - 1, p - w);
      if ((y > 1) && (y - 1 != x)) upd(y - 1, k - t);
    }
  }
}

```