#include <bits/stdc++.h>
using namespace std;
inline int ___INT() {
  int ret;
  scanf("%d", &ret);
  return ret;
}
int N;
int _P[300005];
int P[300005];
int BIT[300005];
void update(int pos, int add) {
  while (pos <= N) {
    BIT[pos] += add;
    pos += pos & -pos;
  }
}
int getVal(int pos) { return int(pos > 1 && P[pos] < P[pos - 1]); }
void init_bit() {
  for (int i = (1); i <= (N); ++i) {
    update(i, getVal(i));
  }
}
int getSum(int id) {
  int ret = 0;
  while (id > 0) {
    ret += BIT[id];
    id -= id & -id;
  }
  return ret;
}
int ask(int u, int v) {
  int S = getSum(v);
  S -= getSum(u - 1);
  if (u > 1 && P[u] < P[u - 1]) --S;
  return S;
}
void doit(int u, int v) {
  int _u = u, _v = v;
  u = _P[u];
  v = _P[v];
  swap(_P[_u], _P[_v]);
  if (u == v) return;
  set<int> critical;
  critical.insert(u);
  if (u + 1 <= N) critical.insert(u + 1);
  critical.insert(v);
  if (v + 1 <= N) critical.insert(v + 1);
  vector<int> old;
  for (set<int>::iterator it = critical.begin(); it != critical.end(); ++it) {
    old.push_back(getVal(*it));
  }
  swap(P[u], P[v]);
  vector<int> New;
  for (set<int>::iterator it = critical.begin(); it != critical.end(); ++it) {
    New.push_back(getVal(*it));
  }
  int k = 0;
  for (set<int>::iterator it = critical.begin(); it != critical.end(); ++it) {
    update(*it, New[k] - old[k]);
    ++k;
  }
}
int main() {
  N = ___INT();
  for (int i = (1); i <= (N); ++i) {
    int x = ___INT();
    _P[i] = x;
    P[x] = i;
  }
  init_bit();
  int q = ___INT();
  while (q--) {
    int t = ___INT();
    int u = ___INT();
    int v = ___INT();
    if (t == 1) {
      printf("%d\n", 1 + ask(u, v));
    } else {
      doit(u, v);
    }
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
inline int ___INT() {
  int ret;
  scanf("%d", &ret);
  return ret;
}
int N;
int _P[300005];
int P[300005];
int BIT[300005];
void update(int pos, int add) {
  while (pos <= N) {
    BIT[pos] += add;
    pos += pos & -pos;
  }
}
int getVal(int pos) { return int(pos > 1 && P[pos] < P[pos - 1]); }
void init_bit() {
  for (int i = (1); i <= (N); ++i) {
    update(i, getVal(i));
  }
}
int getSum(int id) {
  int ret = 0;
  while (id > 0) {
    ret += BIT[id];
    id -= id & -id;
  }
  return ret;
}
int ask(int u, int v) {
  int S = getSum(v);
  S -= getSum(u - 1);
  if (u > 1 && P[u] < P[u - 1]) --S;
  return S;
}
void doit(int u, int v) {
  int _u = u, _v = v;
  u = _P[u];
  v = _P[v];
  swap(_P[_u], _P[_v]);
  if (u == v) return;
  set<int> critical;
  critical.insert(u);
  if (u + 1 <= N) critical.insert(u + 1);
  critical.insert(v);
  if (v + 1 <= N) critical.insert(v + 1);
  vector<int> old;
  for (set<int>::iterator it = critical.begin(); it != critical.end(); ++it) {
    old.push_back(getVal(*it));
  }
  swap(P[u], P[v]);
  vector<int> New;
  for (set<int>::iterator it = critical.begin(); it != critical.end(); ++it) {
    New.push_back(getVal(*it));
  }
  int k = 0;
  for (set<int>::iterator it = critical.begin(); it != critical.end(); ++it) {
    update(*it, New[k] - old[k]);
    ++k;
  }
}
int main() {
  N = ___INT();
  for (int i = (1); i <= (N); ++i) {
    int x = ___INT();
    _P[i] = x;
    P[x] = i;
  }
  init_bit();
  int q = ___INT();
  while (q--) {
    int t = ___INT();
    int u = ___INT();
    int v = ___INT();
    if (t == 1) {
      printf("%d\n", 1 + ask(u, v));
    } else {
      doit(u, v);
    }
  }
  return 0;
}

```