#include <bits/stdc++.h>
using namespace std;
inline void pisz(int n) { printf("%d\n", n); }
template <typename T, typename TT>
ostream &operator<<(ostream &s, pair<T, TT> t) {
  return s << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream &operator<<(ostream &s, vector<T> t) {
  for (int(i) = 0; (i) < (((int)((t).size()))); ++(i)) s << t[i] << " ";
  return s;
}
void add(int *a, int n, int x) {
  for (; n < 400007; n |= n + 1) a[n] += x;
}
int sum(int *a, int n) {
  int s = 0;
  while (n >= 0) {
    s += a[n];
    n = (n & (n + 1)) - 1;
  }
  return s;
}
int sumbetween(int *a, int le, int ri) { return sum(a, ri) - sum(a, le - 1); }
int fen[400007], a[400007], pos[400007], val[400007];
void upd(int i) {
  if (pos[i + 1] < pos[i]) {
    if (val[i] != 1) {
      val[i] = 1;
      add(fen, i, 1);
    }
  } else {
    if (val[i] != 0) {
      val[i] = 0;
      add(fen, i, -1);
    }
  }
}
int main() {
  int(n);
  scanf("%d", &(n));
  for (int(i) = 1; (i) <= (n); ++(i)) {
    scanf("%d", a + i);
    pos[a[i]] = i;
  }
  for (int(i) = 1; (i) <= (n - 1); ++(i))
    if (pos[i + 1] < pos[i]) {
      val[i] = 1;
      add(fen, i, 1);
    }
  int(testow);
  scanf("%d", &(testow));
  while (testow--) {
    int(typ), (x), (y);
    scanf("%d %d %d", &(typ), &(x), &(y));
    if (typ == 1) {
      pisz(1 + sumbetween(fen, x, y - 1));
    } else {
      int A = a[x], B = a[y];
      pos[A] = y;
      pos[B] = x;
      a[x] = B;
      a[y] = A;
      upd(A);
      if (A > 1) upd(A - 1);
      upd(B);
      if (B > 1) upd(B - 1);
    }
  }
}

### 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;
inline void pisz(int n) { printf("%d\n", n); }
template <typename T, typename TT>
ostream &operator<<(ostream &s, pair<T, TT> t) {
  return s << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream &operator<<(ostream &s, vector<T> t) {
  for (int(i) = 0; (i) < (((int)((t).size()))); ++(i)) s << t[i] << " ";
  return s;
}
void add(int *a, int n, int x) {
  for (; n < 400007; n |= n + 1) a[n] += x;
}
int sum(int *a, int n) {
  int s = 0;
  while (n >= 0) {
    s += a[n];
    n = (n & (n + 1)) - 1;
  }
  return s;
}
int sumbetween(int *a, int le, int ri) { return sum(a, ri) - sum(a, le - 1); }
int fen[400007], a[400007], pos[400007], val[400007];
void upd(int i) {
  if (pos[i + 1] < pos[i]) {
    if (val[i] != 1) {
      val[i] = 1;
      add(fen, i, 1);
    }
  } else {
    if (val[i] != 0) {
      val[i] = 0;
      add(fen, i, -1);
    }
  }
}
int main() {
  int(n);
  scanf("%d", &(n));
  for (int(i) = 1; (i) <= (n); ++(i)) {
    scanf("%d", a + i);
    pos[a[i]] = i;
  }
  for (int(i) = 1; (i) <= (n - 1); ++(i))
    if (pos[i + 1] < pos[i]) {
      val[i] = 1;
      add(fen, i, 1);
    }
  int(testow);
  scanf("%d", &(testow));
  while (testow--) {
    int(typ), (x), (y);
    scanf("%d %d %d", &(typ), &(x), &(y));
    if (typ == 1) {
      pisz(1 + sumbetween(fen, x, y - 1));
    } else {
      int A = a[x], B = a[y];
      pos[A] = y;
      pos[B] = x;
      a[x] = B;
      a[y] = A;
      upd(A);
      if (A > 1) upd(A - 1);
      upd(B);
      if (B > 1) upd(B - 1);
    }
  }
}

```