#include <bits/stdc++.h>
using namespace std;
int t[666666], a[333333], p[333333], n;
int sum(int r) {
  int result = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) result += t[r];
  return result;
}
void inc(int i, int delta) {
  for (; i < n + 5; i = (i | (i + 1))) t[i] += delta;
}
int sum(int l, int r) { return sum(r) - sum(l) + 1; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    p[a[i]] = i;
  }
  p[n + 1] = n + 1;
  for (int i = 1; i <= n; i++)
    if (p[i] < p[i - 1]) inc(i, 1);
  int q;
  cin >> q;
  for (int t = 1; t <= q; t++) {
    int pi, xi, yi, xi1, yi1;
    cin >> pi >> xi1 >> yi1;
    if (pi == 1) {
      cout << sum(xi1, yi1) << endl;
    } else {
      int xx = a[xi1];
      int yy = a[yi1];
      if (xx < n) {
        int nxt = p[xx + 1];
        bool curr = nxt < p[xx];
        if (curr) inc(xx + 1, -1);
        bool became = nxt < (yi1);
        if (became) inc(xx + 1, 1);
      }
      if (xx > 1) {
        int prv = p[xx - 1];
        bool curr = p[xx] < prv;
        if (curr) inc(xx, -1);
        bool became = (yi1) < prv;
        if (became) inc(xx, 1);
      }
      p[xx] = yi1;
      if (yy < n) {
        int nxt = p[yy + 1];
        bool curr = nxt < p[yy];
        if (curr) inc(yy + 1, -1);
        bool became = nxt < (xi1);
        if (became) inc(yy + 1, 1);
      }
      if (yy > 1) {
        int prv = p[yy - 1];
        bool curr = p[yy] < prv;
        if (curr) inc(yy, -1);
        bool became = (xi1) < prv;
        if (became) inc(yy, 1);
      }
      p[yy] = xi1;
      swap(a[xi1], a[yi1]);
    }
  }
  cin >> n;
  return 0;
}

### Prompt

Create a solution in Cpp for 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 t[666666], a[333333], p[333333], n;
int sum(int r) {
  int result = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) result += t[r];
  return result;
}
void inc(int i, int delta) {
  for (; i < n + 5; i = (i | (i + 1))) t[i] += delta;
}
int sum(int l, int r) { return sum(r) - sum(l) + 1; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    p[a[i]] = i;
  }
  p[n + 1] = n + 1;
  for (int i = 1; i <= n; i++)
    if (p[i] < p[i - 1]) inc(i, 1);
  int q;
  cin >> q;
  for (int t = 1; t <= q; t++) {
    int pi, xi, yi, xi1, yi1;
    cin >> pi >> xi1 >> yi1;
    if (pi == 1) {
      cout << sum(xi1, yi1) << endl;
    } else {
      int xx = a[xi1];
      int yy = a[yi1];
      if (xx < n) {
        int nxt = p[xx + 1];
        bool curr = nxt < p[xx];
        if (curr) inc(xx + 1, -1);
        bool became = nxt < (yi1);
        if (became) inc(xx + 1, 1);
      }
      if (xx > 1) {
        int prv = p[xx - 1];
        bool curr = p[xx] < prv;
        if (curr) inc(xx, -1);
        bool became = (yi1) < prv;
        if (became) inc(xx, 1);
      }
      p[xx] = yi1;
      if (yy < n) {
        int nxt = p[yy + 1];
        bool curr = nxt < p[yy];
        if (curr) inc(yy + 1, -1);
        bool became = nxt < (xi1);
        if (became) inc(yy + 1, 1);
      }
      if (yy > 1) {
        int prv = p[yy - 1];
        bool curr = p[yy] < prv;
        if (curr) inc(yy, -1);
        bool became = (xi1) < prv;
        if (became) inc(yy, 1);
      }
      p[yy] = xi1;
      swap(a[xi1], a[yi1]);
    }
  }
  cin >> n;
  return 0;
}

```