#include <bits/stdc++.h>
using namespace std;
const int dy[] = {-1, 0, 1, 0}, dx[] = {0, 1, 0, -1};
const double EPS = 1e-8;
const double PI = acos(-1.0);
int popcount(int n) { return __builtin_popcount(n); }
int popcount(long long n) { return __builtin_popcountll(n); }
template <class T>
int SIZE(T a) {
  return a.size();
}
template <class T>
string IntToString(T num) {
  string res;
  stringstream ss;
  ss << num;
  return ss.str();
}
template <class T>
T StringToInt(string str) {
  T res = 0;
  for (int i = 0; i < SIZE(str); i++) res = (res * 10 + str[i] - '0');
  return res;
}
template <class T>
T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
void PrintSeq(T &a, int sz) {
  for (int i = 0; i < sz; i++) {
    cout << a[i];
    if (sz == i + 1)
      cout << endl;
    else
      cout << ' ';
  }
}
long long getTen(int a) { return (a <= 0) ? 1 : (getTen(a - 1) * 10); }
bool EQ(double a, double b) { return abs(a - b) < EPS; }
void fastStream() {
  cin.tie(0);
  std::ios_base::sync_with_stdio(0);
}
vector<string> split(string str, char del) {
  vector<string> res;
  for (int i = 0, s = 0; i < SIZE(str); i++) {
    if (str[i] == del) {
      if (i - s != 0) res.push_back(str.substr(s, i - s));
      s = i + 1;
    } else if (i == SIZE(str) - 1) {
      res.push_back(str.substr(s));
    }
  }
  return res;
}
int N;
int as[500001];
int valToPos[500010];
int posToVal[500010];
const int MAX_N = 5000100;
class BIT {
 private:
  long long bit[MAX_N];
  int n;

 public:
  BIT(int sz) {
    n = sz;
    for (int i = 0; i <= n; i++) bit[i] = 0;
  }
  long long sum(int i) {
    long long s = 0;
    while (i > 0) {
      s += bit[i];
      i -= i & -i;
    }
    return s;
  }
  long long rangeSum(int from, int to) {
    long long res = 0;
    res = sum(to) - sum(from - 1);
    return res;
  }
  void add(int i, long long x) {
    while (i <= n) {
      bit[i] += x;
      i += i & -i;
    }
  }
  void insert(int i, long long x) { add(i, -(sum(i) - sum(i - 1)) + x); }
};
BIT bit(1000000);
int main() {
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> as[i];
    posToVal[i] = as[i];
    valToPos[as[i]] = i;
  }
  for (int i = 0; i < N; i++) {
    if (as[i] == 1)
      bit.add(1, 0);
    else {
      int pos1 = i;
      int pos2 = valToPos[as[i] - 1];
      if (pos2 > pos1) {
        bit.add(as[i], 1);
      }
    }
  }
  int Q;
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    int p, x, y;
    cin >> p >> x >> y;
    if (p == 1) {
      cout << bit.rangeSum(x, y) + 1 - bit.rangeSum(x, x) << endl;
    } else {
      int posx = x - 1;
      int posy = y - 1;
      x = posToVal[posx];
      y = posToVal[posy];
      valToPos[x] = posy;
      valToPos[y] = posx;
      posToVal[posx] = y;
      posToVal[posy] = x;
      int ary[6] = {x, x - 1, x + 1, y, y - 1, y + 1};
      for (int j = 0; j < 6; j++) {
        int num = ary[j];
        if (num == 0)
          continue;
        else if (num == 1)
          continue;
        else {
          int pos1 = valToPos[num];
          int pos2 = valToPos[num - 1];
          if (pos2 > pos1)
            bit.insert(num, 1);
          else
            bit.insert(num, 0);
        }
      }
    }
  }
  return 0;
}

### Prompt

Generate 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;
const int dy[] = {-1, 0, 1, 0}, dx[] = {0, 1, 0, -1};
const double EPS = 1e-8;
const double PI = acos(-1.0);
int popcount(int n) { return __builtin_popcount(n); }
int popcount(long long n) { return __builtin_popcountll(n); }
template <class T>
int SIZE(T a) {
  return a.size();
}
template <class T>
string IntToString(T num) {
  string res;
  stringstream ss;
  ss << num;
  return ss.str();
}
template <class T>
T StringToInt(string str) {
  T res = 0;
  for (int i = 0; i < SIZE(str); i++) res = (res * 10 + str[i] - '0');
  return res;
}
template <class T>
T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
void PrintSeq(T &a, int sz) {
  for (int i = 0; i < sz; i++) {
    cout << a[i];
    if (sz == i + 1)
      cout << endl;
    else
      cout << ' ';
  }
}
long long getTen(int a) { return (a <= 0) ? 1 : (getTen(a - 1) * 10); }
bool EQ(double a, double b) { return abs(a - b) < EPS; }
void fastStream() {
  cin.tie(0);
  std::ios_base::sync_with_stdio(0);
}
vector<string> split(string str, char del) {
  vector<string> res;
  for (int i = 0, s = 0; i < SIZE(str); i++) {
    if (str[i] == del) {
      if (i - s != 0) res.push_back(str.substr(s, i - s));
      s = i + 1;
    } else if (i == SIZE(str) - 1) {
      res.push_back(str.substr(s));
    }
  }
  return res;
}
int N;
int as[500001];
int valToPos[500010];
int posToVal[500010];
const int MAX_N = 5000100;
class BIT {
 private:
  long long bit[MAX_N];
  int n;

 public:
  BIT(int sz) {
    n = sz;
    for (int i = 0; i <= n; i++) bit[i] = 0;
  }
  long long sum(int i) {
    long long s = 0;
    while (i > 0) {
      s += bit[i];
      i -= i & -i;
    }
    return s;
  }
  long long rangeSum(int from, int to) {
    long long res = 0;
    res = sum(to) - sum(from - 1);
    return res;
  }
  void add(int i, long long x) {
    while (i <= n) {
      bit[i] += x;
      i += i & -i;
    }
  }
  void insert(int i, long long x) { add(i, -(sum(i) - sum(i - 1)) + x); }
};
BIT bit(1000000);
int main() {
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> as[i];
    posToVal[i] = as[i];
    valToPos[as[i]] = i;
  }
  for (int i = 0; i < N; i++) {
    if (as[i] == 1)
      bit.add(1, 0);
    else {
      int pos1 = i;
      int pos2 = valToPos[as[i] - 1];
      if (pos2 > pos1) {
        bit.add(as[i], 1);
      }
    }
  }
  int Q;
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    int p, x, y;
    cin >> p >> x >> y;
    if (p == 1) {
      cout << bit.rangeSum(x, y) + 1 - bit.rangeSum(x, x) << endl;
    } else {
      int posx = x - 1;
      int posy = y - 1;
      x = posToVal[posx];
      y = posToVal[posy];
      valToPos[x] = posy;
      valToPos[y] = posx;
      posToVal[posx] = y;
      posToVal[posy] = x;
      int ary[6] = {x, x - 1, x + 1, y, y - 1, y + 1};
      for (int j = 0; j < 6; j++) {
        int num = ary[j];
        if (num == 0)
          continue;
        else if (num == 1)
          continue;
        else {
          int pos1 = valToPos[num];
          int pos2 = valToPos[num - 1];
          if (pos2 > pos1)
            bit.insert(num, 1);
          else
            bit.insert(num, 0);
        }
      }
    }
  }
  return 0;
}

```