#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int s = 0;
  bool f = 0;
  char ch = ' ';
  while (!isdigit(ch)) {
    f |= (ch == '-');
    ch = getchar();
  }
  while (isdigit(ch)) {
    s = (s << 3) + (s << 1) + (ch ^ 48);
    ch = getchar();
  }
  return (f) ? (-s) : (s);
}
inline void write(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x < 10) {
    putchar(x + '0');
    return;
  }
  write(x / 10);
  putchar((x % 10) + '0');
}
const int N = 300005;
int n, Q, a[N], pos[N];
struct Node {
  int sum;
} T[N << 2];
inline Node merg(Node p, Node q) {
  Node ans;
  ans.sum = p.sum + q.sum;
  return ans;
}
inline void chag(int x, int l, int r, int pos, int val) {
  if (l == r) {
    T[x].sum = val;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    chag((x << 1), l, mid, pos, val);
  else
    chag((x << 1 | 1), mid + 1, r, pos, val);
  T[x] = merg(T[(x << 1)], T[(x << 1 | 1)]);
}
inline Node que(int x, int l, int r, int ql, int qr) {
  if (l == ql && r == qr) return T[x];
  int mid = (l + r) >> 1;
  if (qr <= mid)
    return que((x << 1), l, mid, ql, qr);
  else if (ql > mid)
    return que((x << 1 | 1), mid + 1, r, ql, qr);
  else
    return merg(que((x << 1), l, mid, ql, mid),
                que((x << 1 | 1), mid + 1, r, mid + 1, qr));
  T[x] = merg(T[(x << 1)], T[(x << 1 | 1)]);
}
int main() {
  int i, opt, x, y;
  n = read();
  for (i = 1; i <= n; i++) pos[a[i] = read()] = i;
  for (i = 1; i < n; i++)
    if (pos[i] > pos[i + 1]) chag(1, 1, n, i, 1);
  Q = read();
  while (Q--) {
    opt = read();
    x = read();
    y = read();
    if (opt == 1) {
      if (x == y)
        write(1), putchar('\n');
      else
        write(que(1, 1, n, x, y - 1).sum + 1), putchar('\n');
    } else {
      int xx = a[x], yy = a[y];
      swap(pos[a[x]], pos[a[y]]);
      swap(a[x], a[y]);
      if (xx != n && pos[xx] > pos[xx + 1]) chag(1, 1, n, xx, 1);
      if (xx != n && pos[xx] < pos[xx + 1]) chag(1, 1, n, xx, 0);
      if (xx != 1 && pos[xx - 1] > pos[xx]) chag(1, 1, n, xx - 1, 1);
      if (xx != 1 && pos[xx - 1] < pos[xx]) chag(1, 1, n, xx - 1, 0);
      if (yy != n && pos[yy] > pos[yy + 1]) chag(1, 1, n, yy, 1);
      if (yy != n && pos[yy] < pos[yy + 1]) chag(1, 1, n, yy, 0);
      if (yy != 1 && pos[yy - 1] > pos[yy]) chag(1, 1, n, yy - 1, 1);
      if (yy != 1 && pos[yy - 1] < pos[yy]) chag(1, 1, n, yy - 1, 0);
    }
  }
  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;
inline int read() {
  int s = 0;
  bool f = 0;
  char ch = ' ';
  while (!isdigit(ch)) {
    f |= (ch == '-');
    ch = getchar();
  }
  while (isdigit(ch)) {
    s = (s << 3) + (s << 1) + (ch ^ 48);
    ch = getchar();
  }
  return (f) ? (-s) : (s);
}
inline void write(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x < 10) {
    putchar(x + '0');
    return;
  }
  write(x / 10);
  putchar((x % 10) + '0');
}
const int N = 300005;
int n, Q, a[N], pos[N];
struct Node {
  int sum;
} T[N << 2];
inline Node merg(Node p, Node q) {
  Node ans;
  ans.sum = p.sum + q.sum;
  return ans;
}
inline void chag(int x, int l, int r, int pos, int val) {
  if (l == r) {
    T[x].sum = val;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    chag((x << 1), l, mid, pos, val);
  else
    chag((x << 1 | 1), mid + 1, r, pos, val);
  T[x] = merg(T[(x << 1)], T[(x << 1 | 1)]);
}
inline Node que(int x, int l, int r, int ql, int qr) {
  if (l == ql && r == qr) return T[x];
  int mid = (l + r) >> 1;
  if (qr <= mid)
    return que((x << 1), l, mid, ql, qr);
  else if (ql > mid)
    return que((x << 1 | 1), mid + 1, r, ql, qr);
  else
    return merg(que((x << 1), l, mid, ql, mid),
                que((x << 1 | 1), mid + 1, r, mid + 1, qr));
  T[x] = merg(T[(x << 1)], T[(x << 1 | 1)]);
}
int main() {
  int i, opt, x, y;
  n = read();
  for (i = 1; i <= n; i++) pos[a[i] = read()] = i;
  for (i = 1; i < n; i++)
    if (pos[i] > pos[i + 1]) chag(1, 1, n, i, 1);
  Q = read();
  while (Q--) {
    opt = read();
    x = read();
    y = read();
    if (opt == 1) {
      if (x == y)
        write(1), putchar('\n');
      else
        write(que(1, 1, n, x, y - 1).sum + 1), putchar('\n');
    } else {
      int xx = a[x], yy = a[y];
      swap(pos[a[x]], pos[a[y]]);
      swap(a[x], a[y]);
      if (xx != n && pos[xx] > pos[xx + 1]) chag(1, 1, n, xx, 1);
      if (xx != n && pos[xx] < pos[xx + 1]) chag(1, 1, n, xx, 0);
      if (xx != 1 && pos[xx - 1] > pos[xx]) chag(1, 1, n, xx - 1, 1);
      if (xx != 1 && pos[xx - 1] < pos[xx]) chag(1, 1, n, xx - 1, 0);
      if (yy != n && pos[yy] > pos[yy + 1]) chag(1, 1, n, yy, 1);
      if (yy != n && pos[yy] < pos[yy + 1]) chag(1, 1, n, yy, 0);
      if (yy != 1 && pos[yy - 1] > pos[yy]) chag(1, 1, n, yy - 1, 1);
      if (yy != 1 && pos[yy - 1] < pos[yy]) chag(1, 1, n, yy - 1, 0);
    }
  }
  return 0;
}

```