#include <bits/stdc++.h>
using namespace std;
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) {
  s << "{";
  for (int i = 0; i < t.size(); i++)
    s << t[i] << (i == t.size() - 1 ? "" : ",");
  return s << "}" << endl;
}
inline void pisz(int x) { printf("%d\n", x); }
int bit[1000008];
int maxval;
void czysc(int mxv) {
  maxval = mxv;
  for (int i = 0; i < mxv + 1; i++) bit[i] = 0;
}
void dodaj(int ind, int war) {
  ind++;
  while (ind < maxval) {
    bit[ind] += war;
    ind += (ind & -ind);
  }
}
int suma(int ind) {
  ind++;
  int sm = 0;
  while (ind > 0) {
    sm += bit[ind];
    ind -= (ind & -ind);
  }
  return sm;
}
int suma(int ind1, int ind2) { return suma(ind2) - suma(ind1 - 1); }
int a[1 << 20];
int pos[1 << 20];
int old[1 << 20];
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  scanf("%d", &(n));
  ;
  for (int i = 0; i < n; i++) {
    scanf("%d", &(a[i]));
    ;
    a[i]--;
    pos[a[i]] = i;
  }
  czysc(n + 2);
  for (int i = 0; i < n - 1; i++) {
    if (pos[i] > pos[i + 1]) {
      old[i] = 1;
      dodaj(i, 1);
    }
  }
  int q;
  scanf("%d", &(q));
  ;
  while (q--) {
    int t;
    scanf("%d", &(t));
    ;
    if (t == 1) {
      int x, y;
      scanf("%d%d", &(x), &(y));
      ;
      x--;
      y--;
      printf("%d\n", suma(x, y - 1) + 1);
    } else {
      int x, y;
      scanf("%d%d", &(x), &(y));
      ;
      x--;
      y--;
      swap(a[x], a[y]);
      pos[a[x]] = x;
      pos[a[y]] = y;
      x = a[x];
      y = a[y];
      for (int k = x - 1; k <= x + 1; k++) {
        if (0 <= k && k < n - 1) {
          dodaj(k, -old[k]);
          if (pos[k] > pos[k + 1]) {
            old[k] = 1;
            dodaj(k, 1);
          } else
            old[k] = 0;
        }
      }
      for (int k = y - 1; k <= y + 1; k++) {
        if (0 <= k && k < n - 1) {
          dodaj(k, -old[k]);
          if (pos[k] > pos[k + 1]) {
            old[k] = 1;
            dodaj(k, 1);
          } else
            old[k] = 0;
        }
      }
    }
  }
  return 0;
}

### Prompt

Your task is to create 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;
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) {
  s << "{";
  for (int i = 0; i < t.size(); i++)
    s << t[i] << (i == t.size() - 1 ? "" : ",");
  return s << "}" << endl;
}
inline void pisz(int x) { printf("%d\n", x); }
int bit[1000008];
int maxval;
void czysc(int mxv) {
  maxval = mxv;
  for (int i = 0; i < mxv + 1; i++) bit[i] = 0;
}
void dodaj(int ind, int war) {
  ind++;
  while (ind < maxval) {
    bit[ind] += war;
    ind += (ind & -ind);
  }
}
int suma(int ind) {
  ind++;
  int sm = 0;
  while (ind > 0) {
    sm += bit[ind];
    ind -= (ind & -ind);
  }
  return sm;
}
int suma(int ind1, int ind2) { return suma(ind2) - suma(ind1 - 1); }
int a[1 << 20];
int pos[1 << 20];
int old[1 << 20];
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  scanf("%d", &(n));
  ;
  for (int i = 0; i < n; i++) {
    scanf("%d", &(a[i]));
    ;
    a[i]--;
    pos[a[i]] = i;
  }
  czysc(n + 2);
  for (int i = 0; i < n - 1; i++) {
    if (pos[i] > pos[i + 1]) {
      old[i] = 1;
      dodaj(i, 1);
    }
  }
  int q;
  scanf("%d", &(q));
  ;
  while (q--) {
    int t;
    scanf("%d", &(t));
    ;
    if (t == 1) {
      int x, y;
      scanf("%d%d", &(x), &(y));
      ;
      x--;
      y--;
      printf("%d\n", suma(x, y - 1) + 1);
    } else {
      int x, y;
      scanf("%d%d", &(x), &(y));
      ;
      x--;
      y--;
      swap(a[x], a[y]);
      pos[a[x]] = x;
      pos[a[y]] = y;
      x = a[x];
      y = a[y];
      for (int k = x - 1; k <= x + 1; k++) {
        if (0 <= k && k < n - 1) {
          dodaj(k, -old[k]);
          if (pos[k] > pos[k + 1]) {
            old[k] = 1;
            dodaj(k, 1);
          } else
            old[k] = 0;
        }
      }
      for (int k = y - 1; k <= y + 1; k++) {
        if (0 <= k && k < n - 1) {
          dodaj(k, -old[k]);
          if (pos[k] > pos[k + 1]) {
            old[k] = 1;
            dodaj(k, 1);
          } else
            old[k] = 0;
        }
      }
    }
  }
  return 0;
}

```