#include <bits/stdc++.h>
using namespace std;
bool debug = false;
int n, m, k;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
long long ln, lk, lm;
int a[1005], pos[1005], T[1005], M;
void update(int x, int v) {
  x += M;
  T[x] = v;
  while (x > 1) {
    T[x >> 1] = T[x] + T[x ^ 1];
    x >>= 1;
  }
}
int qy(int l, int r) {
  if (r < l) return 0;
  int ans = 0;
  for (l = l + M - 1, r = r + M + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
    if (~l & 1) ans += T[l ^ 1];
    if (r & 1) ans += T[r ^ 1];
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (M = 1; n + 2 > M; M <<= 1)
    ;
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    scanf("%d", a + i);
    pos[a[i]] = i;
  }
  for (int(i) = 1; (i) <= (int)(n - 1); (i)++) {
    if (pos[i] > pos[i + 1]) update(i, 1);
  }
  int q, t, x, y;
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d", &t, &x, &y);
    if (t == 1) {
      printf("%d\n", qy(x, y - 1) + 1);
    } else if (t == 2) {
      int ida = a[x], idy = a[y];
      if (ida < n) update(ida, 0);
      if (idy < n) update(idy, 0);
      if (ida > 1) update(ida - 1, 0);
      if (idy > 1) update(idy - 1, 0);
      pos[ida] = y;
      pos[idy] = x;
      swap(a[x], a[y]);
      if (ida < n) {
        if (pos[ida] > pos[ida + 1]) update(ida, 1);
      }
      if (idy < n && pos[idy] > pos[idy + 1]) update(idy, 1);
      if (ida > 1 && pos[ida] < pos[ida - 1]) update(ida - 1, 1);
      if (idy > 1 && pos[idy] < pos[idy - 1]) update(idy - 1, 1);
    }
  }
  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>
using namespace std;
bool debug = false;
int n, m, k;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
long long ln, lk, lm;
int a[1005], pos[1005], T[1005], M;
void update(int x, int v) {
  x += M;
  T[x] = v;
  while (x > 1) {
    T[x >> 1] = T[x] + T[x ^ 1];
    x >>= 1;
  }
}
int qy(int l, int r) {
  if (r < l) return 0;
  int ans = 0;
  for (l = l + M - 1, r = r + M + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
    if (~l & 1) ans += T[l ^ 1];
    if (r & 1) ans += T[r ^ 1];
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (M = 1; n + 2 > M; M <<= 1)
    ;
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    scanf("%d", a + i);
    pos[a[i]] = i;
  }
  for (int(i) = 1; (i) <= (int)(n - 1); (i)++) {
    if (pos[i] > pos[i + 1]) update(i, 1);
  }
  int q, t, x, y;
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d", &t, &x, &y);
    if (t == 1) {
      printf("%d\n", qy(x, y - 1) + 1);
    } else if (t == 2) {
      int ida = a[x], idy = a[y];
      if (ida < n) update(ida, 0);
      if (idy < n) update(idy, 0);
      if (ida > 1) update(ida - 1, 0);
      if (idy > 1) update(idy - 1, 0);
      pos[ida] = y;
      pos[idy] = x;
      swap(a[x], a[y]);
      if (ida < n) {
        if (pos[ida] > pos[ida + 1]) update(ida, 1);
      }
      if (idy < n && pos[idy] > pos[idy + 1]) update(idy, 1);
      if (ida > 1 && pos[ida] < pos[ida - 1]) update(ida - 1, 1);
      if (idy > 1 && pos[idy] < pos[idy - 1]) update(idy - 1, 1);
    }
  }
  return 0;
}

```