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
#include <bits/stdc++.h>
namespace solution {
class SolutionInterface {
 public:
  virtual int run() = 0;

 protected:
  virtual void pre_calc() {}
  virtual bool action() = 0;
  virtual void init(){};
  virtual bool input() { return false; };
  virtual void output(){};
  SolutionInterface() {}

 private:
};
}  // namespace solution
namespace tree {
class BinaryIndexedTree {
 public:
  int n;
  std::vector<int> data;
  BinaryIndexedTree() {}
  BinaryIndexedTree(int n) : n(n) { data = std::vector<int>(n + 1, 0); }
  int sum(int i) {
    i++;
    int s = 0;
    while (i > 0) {
      s += data[i];
      i -= i & -i;
    }
    return s;
  }
  void add(int i, int x) {
    i++;
    while (i <= n) {
      data[i] += x;
      i += i & -i;
    }
  }
};
}  // namespace tree
namespace solution {
class SolutionBase : public SolutionInterface {
 public:
  virtual int run() {
    pre_calc();
    while (action())
      ;
    return 0;
  }
};
}  // namespace solution
namespace solution {}
namespace solution {
using namespace std;
}
namespace solution {
const int SIZE = 3 * 100000 + 11;
const int NONE = -1;
int n;
int A[SIZE];
int q;
int P[SIZE];
int X[SIZE];
int Y[SIZE];
int B[SIZE];
tree::BinaryIndexedTree bit;
int results[SIZE];
}  // namespace solution
namespace solution {
class Solver {
 public:
  void solve() {
    B[0] = std::numeric_limits<int>::min();
    B[n + 1] = std::numeric_limits<int>::max();
    for (int i = 1; i <= n; ++i) B[A[i]] = i;
    for (int i = 1; i <= n; ++i)
      if (B[i] > B[i + 1]) set_value(i, 1);
    for (int i = 0; i < q; ++i) results[i] = query(P[i], X[i], Y[i]);
  }
  int get_value(int x) { return bit.sum(x) - bit.sum(x - 1); }
  void set_value(int x, int new_value) {
    int value = get_value(x);
    bit.add(x, -value);
    bit.add(x, new_value);
  }
  int query(int p, int x, int y) {
    if (p == 1) {
      return query_a(x, y);
    } else if (p == 2) {
      return query_b(x, y);
    }
    return NONE;
  }
  int query_a(int x, int y) { return bit.sum(y - 1) - bit.sum(x - 1) + 1; }
  int query_b(int x, int y) {
    int value_x = A[x];
    int value_y = A[y];
    B[value_x] = y;
    B[value_y] = x;
    swap(A[x], A[y]);
    for (int i = -1; i <= 1; ++i) {
      int n_value_x = value_x + i;
      int n_value_y = value_y + i;
      if (n_value_x > 0 && n_value_x + 1 <= n) {
        set_value(n_value_x, B[n_value_x] > B[n_value_x + 1] ? 1 : 0);
      }
      if (n_value_y > 0 && n_value_y + 1 <= n) {
        set_value(n_value_y, B[n_value_y] > B[n_value_y + 1] ? 1 : 0);
      }
    }
    return NONE;
  }

 private:
};
}  // namespace solution
namespace solution {
class Solution : public SolutionBase {
 public:
 protected:
  virtual bool action() {
    init();
    if (!input()) return false;
    solver.solve();
    output();
    return true;
  }
  void init() { bit = tree::BinaryIndexedTree(SIZE); }
  bool input() {
    if (!(cin >> n)) return false;
    for (int i = 1; i <= n; ++i) cin >> A[i];
    cin >> q;
    for (int i = 0; i < q; ++i) cin >> P[i] >> X[i] >> Y[i];
    return true;
  }
  void output() {
    for (int i = 0; i < q; ++i)
      if (results[i] != NONE) cout << results[i] << endl;
  }

 private:
  Solver solver;
};
}  // namespace solution
int main() { return solution::Solution().run(); }
