#include <bits/stdc++.h>
using namespace std;
template <class T>
void print(vector<T> v) {
  cerr << "[";
  if (v.size()) cerr << v[0];
  for (int i = (1); i < (v.size()); ++i) cerr << ", " << v[i];
  cerr << "]\n";
}
template <class T>
string i2s(T x) {
  ostringstream o;
  o << x;
  return o.str();
}
vector<string> splt(string s, char c = ' ') {
  vector<string> all;
  int p = 0, np;
  while (np = s.find(c, p), np >= 0) {
    if (np != p) all.push_back(s.substr(p, np - p));
    p = np + 1;
  }
  if (p < s.size()) all.push_back(s.substr(p));
  return all;
}
int m[1001000];
int v[1001000];
int n;
int s[1 << 21];
void add(int p, int x) {
  p += 1 << 20;
  while (p) {
    s[p] += x;
    p >>= 1;
  }
}
int sum(int p) {
  if (p < 0) return 0;
  p += 1 << 20;
  int rv = 0;
  while (p) {
    if (p & 1) {
      p >>= 1;
    } else {
      rv += s[p];
      p = (p - 1) >> 1;
    }
  }
  return rv;
}
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &m[i]);
  for (int i = (0); i < (n); ++i) v[--m[i]] = i;
  for (int i = (0); i < (n - 1); ++i)
    if (v[i] > v[i + 1]) add(i, 1);
  int q;
  cin >> q;
  for (int i = (0); i < (q); ++i) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    b--;
    c--;
    if (a == 1) {
      printf("%d\n", sum(c - 1) - sum(b - 1) + 1);
    } else {
      int nb = m[b];
      int nc = m[c];
      swap(m[b], m[c]);
      b = nb;
      c = nc;
      if (b > c) swap(b, c);
      if (b && v[b] < v[b - 1]) add(b - 1, -1);
      if (v[b + 1] < v[b]) add(b, -1);
      if (b + 1 != c && v[c] < v[c - 1]) add(c - 1, -1);
      if (c < n - 1 && v[c + 1] < v[c]) add(c, -1);
      swap(v[b], v[c]);
      if (b && v[b] < v[b - 1]) add(b - 1, +1);
      if (v[b + 1] < v[b]) add(b, +1);
      if (b + 1 != c && v[c] < v[c - 1]) add(c - 1, +1);
      if (c < n - 1 && v[c + 1] < v[c]) add(c, +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;
template <class T>
void print(vector<T> v) {
  cerr << "[";
  if (v.size()) cerr << v[0];
  for (int i = (1); i < (v.size()); ++i) cerr << ", " << v[i];
  cerr << "]\n";
}
template <class T>
string i2s(T x) {
  ostringstream o;
  o << x;
  return o.str();
}
vector<string> splt(string s, char c = ' ') {
  vector<string> all;
  int p = 0, np;
  while (np = s.find(c, p), np >= 0) {
    if (np != p) all.push_back(s.substr(p, np - p));
    p = np + 1;
  }
  if (p < s.size()) all.push_back(s.substr(p));
  return all;
}
int m[1001000];
int v[1001000];
int n;
int s[1 << 21];
void add(int p, int x) {
  p += 1 << 20;
  while (p) {
    s[p] += x;
    p >>= 1;
  }
}
int sum(int p) {
  if (p < 0) return 0;
  p += 1 << 20;
  int rv = 0;
  while (p) {
    if (p & 1) {
      p >>= 1;
    } else {
      rv += s[p];
      p = (p - 1) >> 1;
    }
  }
  return rv;
}
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &m[i]);
  for (int i = (0); i < (n); ++i) v[--m[i]] = i;
  for (int i = (0); i < (n - 1); ++i)
    if (v[i] > v[i + 1]) add(i, 1);
  int q;
  cin >> q;
  for (int i = (0); i < (q); ++i) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    b--;
    c--;
    if (a == 1) {
      printf("%d\n", sum(c - 1) - sum(b - 1) + 1);
    } else {
      int nb = m[b];
      int nc = m[c];
      swap(m[b], m[c]);
      b = nb;
      c = nc;
      if (b > c) swap(b, c);
      if (b && v[b] < v[b - 1]) add(b - 1, -1);
      if (v[b + 1] < v[b]) add(b, -1);
      if (b + 1 != c && v[c] < v[c - 1]) add(c - 1, -1);
      if (c < n - 1 && v[c + 1] < v[c]) add(c, -1);
      swap(v[b], v[c]);
      if (b && v[b] < v[b - 1]) add(b - 1, +1);
      if (v[b + 1] < v[b]) add(b, +1);
      if (b + 1 != c && v[c] < v[c - 1]) add(c - 1, +1);
      if (c < n - 1 && v[c + 1] < v[c]) add(c, +1);
    }
  }
  return 0;
}

```