#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
int n;
int a[N];
int b[N];
int t[N], tt[N];
void ubd(int x, int y) {
  if (!(0 <= tt[x] + y && tt[x] + y <= 1)) return;
  tt[x] += y;
  ++x;
  while (x <= n + 1) {
    t[x] += y;
    x += (x & (-x));
  }
}
int qry(int l, int r) {
  ++r;
  if (l > r) return 0;
  int ans = 0;
  while (r > 0) {
    ans += t[r];
    r -= (r & (-r));
  }
  while (l > 0) {
    ans -= t[l];
    l -= (l & (-l));
  }
  return ans;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    a[x] = i;
    if (a[x - 1] == 0) ubd(x, 1);
    b[i] = x;
  }
  a[0] = N;
  a[n + 1] = N;
  int q;
  cin >> q;
  while (q--) {
    int ty, x, y;
    cin >> ty >> x >> y;
    if (ty == 1) {
      cout << qry(x + 1, y) + 1 << endl;
    } else {
      swap(b[x], b[y]);
      x = b[x];
      y = b[y];
      if (a[x] < a[x - 1]) ubd(x, -1);
      if (a[y] < a[y - 1]) ubd(y, -1);
      if (a[x + 1] < a[x]) ubd(x + 1, -1);
      if (a[y + 1] < a[y]) ubd(y + 1, -1);
      swap(a[x], a[y]);
      if (a[x] < a[x - 1]) ubd(x, 1);
      if (a[y] < a[y - 1]) ubd(y, 1);
      if (a[x + 1] < a[x]) ubd(x + 1, 1);
      if (a[y + 1] < a[y]) ubd(y + 1, 1);
    }
  }
  return 0;
}

### 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;
const int N = 300005;
int n;
int a[N];
int b[N];
int t[N], tt[N];
void ubd(int x, int y) {
  if (!(0 <= tt[x] + y && tt[x] + y <= 1)) return;
  tt[x] += y;
  ++x;
  while (x <= n + 1) {
    t[x] += y;
    x += (x & (-x));
  }
}
int qry(int l, int r) {
  ++r;
  if (l > r) return 0;
  int ans = 0;
  while (r > 0) {
    ans += t[r];
    r -= (r & (-r));
  }
  while (l > 0) {
    ans -= t[l];
    l -= (l & (-l));
  }
  return ans;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    a[x] = i;
    if (a[x - 1] == 0) ubd(x, 1);
    b[i] = x;
  }
  a[0] = N;
  a[n + 1] = N;
  int q;
  cin >> q;
  while (q--) {
    int ty, x, y;
    cin >> ty >> x >> y;
    if (ty == 1) {
      cout << qry(x + 1, y) + 1 << endl;
    } else {
      swap(b[x], b[y]);
      x = b[x];
      y = b[y];
      if (a[x] < a[x - 1]) ubd(x, -1);
      if (a[y] < a[y - 1]) ubd(y, -1);
      if (a[x + 1] < a[x]) ubd(x + 1, -1);
      if (a[y + 1] < a[y]) ubd(y + 1, -1);
      swap(a[x], a[y]);
      if (a[x] < a[x - 1]) ubd(x, 1);
      if (a[y] < a[y - 1]) ubd(y, 1);
      if (a[x + 1] < a[x]) ubd(x + 1, 1);
      if (a[y + 1] < a[y]) ubd(y + 1, 1);
    }
  }
  return 0;
}

```