#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:64000000")
int j, n, l, i, h, t, mn, a[300500], f[300500], q1, q2, p1, p2, k, d[300500], m;
bool b[300500];
void tg() {
  freopen("output.txt", "r", stdin);
  freopen("input.txt", "w", stdout);
}
int sum(int p) {
  if (p < 1) {
    return 0;
  }
  return f[p] + sum((p & (p + 1)) - 1);
}
int sum(int l, int r) { return sum(r) - sum(l); }
void inc(int p) {
  for (; p <= n; p |= (p + 1)) {
    f[p]++;
  }
}
void dec(int p) {
  for (; p <= n; p |= (p + 1)) {
    f[p]--;
  }
}
int main() {
  scanf("%d", &n);
  for (j = 1; j <= n; j++) {
    scanf("%d", &a[j]);
    d[a[j]] = j;
  }
  for (j = 2; j <= n; j++) {
    if (d[j - 1] > d[j]) {
      inc(j);
      b[j] = 1;
    }
  }
  scanf("%d", &m);
  for (j = 1; j <= m; j++) {
    scanf("%d%d%d", &t, &q1, &q2);
    if (t == 2) {
      swap(a[q1], a[q2]);
      swap(d[a[q1]], d[a[q2]]);
      q1 = a[q1];
      q2 = a[q2];
      if (q1) {
        if (d[q1 - 1] > d[q1] && !b[q1]) {
          b[q1] = 1;
          inc(q1);
        }
        if (d[q1 - 1] < d[q1] && b[q1]) {
          b[q1] = 0;
          dec(q1);
        }
      }
      if (q1 < n) {
        if (d[q1] > d[q1 + 1] && !b[q1 + 1]) {
          b[q1 + 1] = 1;
          inc(q1 + 1);
        }
        if (d[q1] < d[q1 + 1] && b[q1 + 1]) {
          b[q1 + 1] = 0;
          dec(q1 + 1);
        }
      }
      if (q2) {
        if (d[q2 - 1] > d[q2] && !b[q2]) {
          b[q2] = 1;
          inc(q2);
        }
        if (d[q2 - 1] < d[q2] && b[q2]) {
          b[q2] = 0;
          dec(q2);
        }
      }
      if (q2 < n) {
        if (d[q2] > d[q2 + 1] && !b[q2 + 1]) {
          b[q2 + 1] = 1;
          inc(q2 + 1);
        }
        if (d[q2] < d[q2 + 1] && b[q2 + 1]) {
          b[q2 + 1] = 0;
          dec(q2 + 1);
        }
      }
    } else {
      if (q1 > q2) {
        swap(q1, q2);
      }
      printf("%d\n", sum(q1, q2) + 1);
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
#pragma comment(linker, "/STACK:64000000")
int j, n, l, i, h, t, mn, a[300500], f[300500], q1, q2, p1, p2, k, d[300500], m;
bool b[300500];
void tg() {
  freopen("output.txt", "r", stdin);
  freopen("input.txt", "w", stdout);
}
int sum(int p) {
  if (p < 1) {
    return 0;
  }
  return f[p] + sum((p & (p + 1)) - 1);
}
int sum(int l, int r) { return sum(r) - sum(l); }
void inc(int p) {
  for (; p <= n; p |= (p + 1)) {
    f[p]++;
  }
}
void dec(int p) {
  for (; p <= n; p |= (p + 1)) {
    f[p]--;
  }
}
int main() {
  scanf("%d", &n);
  for (j = 1; j <= n; j++) {
    scanf("%d", &a[j]);
    d[a[j]] = j;
  }
  for (j = 2; j <= n; j++) {
    if (d[j - 1] > d[j]) {
      inc(j);
      b[j] = 1;
    }
  }
  scanf("%d", &m);
  for (j = 1; j <= m; j++) {
    scanf("%d%d%d", &t, &q1, &q2);
    if (t == 2) {
      swap(a[q1], a[q2]);
      swap(d[a[q1]], d[a[q2]]);
      q1 = a[q1];
      q2 = a[q2];
      if (q1) {
        if (d[q1 - 1] > d[q1] && !b[q1]) {
          b[q1] = 1;
          inc(q1);
        }
        if (d[q1 - 1] < d[q1] && b[q1]) {
          b[q1] = 0;
          dec(q1);
        }
      }
      if (q1 < n) {
        if (d[q1] > d[q1 + 1] && !b[q1 + 1]) {
          b[q1 + 1] = 1;
          inc(q1 + 1);
        }
        if (d[q1] < d[q1 + 1] && b[q1 + 1]) {
          b[q1 + 1] = 0;
          dec(q1 + 1);
        }
      }
      if (q2) {
        if (d[q2 - 1] > d[q2] && !b[q2]) {
          b[q2] = 1;
          inc(q2);
        }
        if (d[q2 - 1] < d[q2] && b[q2]) {
          b[q2] = 0;
          dec(q2);
        }
      }
      if (q2 < n) {
        if (d[q2] > d[q2 + 1] && !b[q2 + 1]) {
          b[q2 + 1] = 1;
          inc(q2 + 1);
        }
        if (d[q2] < d[q2 + 1] && b[q2 + 1]) {
          b[q2 + 1] = 0;
          dec(q2 + 1);
        }
      }
    } else {
      if (q1 > q2) {
        swap(q1, q2);
      }
      printf("%d\n", sum(q1, q2) + 1);
    }
  }
  return 0;
}

```