#include <bits/stdc++.h>
int number, qNumber;
int oArray[300001], posArray[300001], prev[300001];
int C[300001];
int Sum(int x) {
  int sum = 0;
  while (x > 0) {
    sum += C[x];
    x -= (-x) & x;
  }
  return sum;
}
int query(int begin, int end) {
  int i;
  int count = 0;
  return Sum(end) - Sum(begin) + 1;
}
void Updata(int pos, int num) {
  while (pos <= number) {
    C[pos] += num;
    pos += (-pos) & pos;
  }
}
void adjust(int begin) {
  if (oArray[begin] != 1) {
    if (posArray[oArray[begin] - 1] < posArray[oArray[begin]]) {
      Updata(oArray[begin], 0 - prev[oArray[begin]]);
      prev[oArray[begin]] = 0;
    } else {
      Updata(oArray[begin], 1 - prev[oArray[begin]]);
      prev[oArray[begin]] = 1;
    }
  }
  if (oArray[begin] != number) {
    if (posArray[oArray[begin]] < posArray[oArray[begin] + 1]) {
      Updata(oArray[begin] + 1, 0 - prev[oArray[begin] + 1]);
      prev[oArray[begin] + 1] = 0;
    } else {
      Updata(oArray[begin] + 1, 1 - prev[oArray[begin] + 1]);
      prev[oArray[begin] + 1] = 1;
    }
  }
}
void swap(int begin, int end) {
  int temp = oArray[begin];
  oArray[begin] = oArray[end];
  oArray[end] = temp;
  posArray[oArray[begin]] = begin;
  posArray[oArray[end]] = end;
  adjust(begin);
  adjust(end);
}
void input() {
  int i, qSign, bPos, ePos;
  for (i = 0; i < number; ++i) {
    scanf("%d", &oArray[i + 1]);
    posArray[oArray[i + 1]] = i + 1;
  }
  memset(C, 0, sizeof(C));
  prev[1] = 1;
  Updata(1, 1);
  for (i = 2; i <= number; ++i) {
    if (posArray[i - 1] < posArray[i])
      prev[i] = 0;
    else {
      prev[i] = 1;
      Updata(i, 1);
    }
  }
  scanf("%d", &qNumber);
  for (i = 0; i < qNumber; ++i) {
    scanf("%d%d%d", &qSign, &bPos, &ePos);
    if (qSign == 1)
      printf("%d\n", query(bPos, ePos));
    else
      swap(bPos, ePos);
  }
}
int main() {
  while (scanf("%d", &number) != EOF) {
    input();
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
int number, qNumber;
int oArray[300001], posArray[300001], prev[300001];
int C[300001];
int Sum(int x) {
  int sum = 0;
  while (x > 0) {
    sum += C[x];
    x -= (-x) & x;
  }
  return sum;
}
int query(int begin, int end) {
  int i;
  int count = 0;
  return Sum(end) - Sum(begin) + 1;
}
void Updata(int pos, int num) {
  while (pos <= number) {
    C[pos] += num;
    pos += (-pos) & pos;
  }
}
void adjust(int begin) {
  if (oArray[begin] != 1) {
    if (posArray[oArray[begin] - 1] < posArray[oArray[begin]]) {
      Updata(oArray[begin], 0 - prev[oArray[begin]]);
      prev[oArray[begin]] = 0;
    } else {
      Updata(oArray[begin], 1 - prev[oArray[begin]]);
      prev[oArray[begin]] = 1;
    }
  }
  if (oArray[begin] != number) {
    if (posArray[oArray[begin]] < posArray[oArray[begin] + 1]) {
      Updata(oArray[begin] + 1, 0 - prev[oArray[begin] + 1]);
      prev[oArray[begin] + 1] = 0;
    } else {
      Updata(oArray[begin] + 1, 1 - prev[oArray[begin] + 1]);
      prev[oArray[begin] + 1] = 1;
    }
  }
}
void swap(int begin, int end) {
  int temp = oArray[begin];
  oArray[begin] = oArray[end];
  oArray[end] = temp;
  posArray[oArray[begin]] = begin;
  posArray[oArray[end]] = end;
  adjust(begin);
  adjust(end);
}
void input() {
  int i, qSign, bPos, ePos;
  for (i = 0; i < number; ++i) {
    scanf("%d", &oArray[i + 1]);
    posArray[oArray[i + 1]] = i + 1;
  }
  memset(C, 0, sizeof(C));
  prev[1] = 1;
  Updata(1, 1);
  for (i = 2; i <= number; ++i) {
    if (posArray[i - 1] < posArray[i])
      prev[i] = 0;
    else {
      prev[i] = 1;
      Updata(i, 1);
    }
  }
  scanf("%d", &qNumber);
  for (i = 0; i < qNumber; ++i) {
    scanf("%d%d%d", &qSign, &bPos, &ePos);
    if (qSign == 1)
      printf("%d\n", query(bPos, ePos));
    else
      swap(bPos, ePos);
  }
}
int main() {
  while (scanf("%d", &number) != EOF) {
    input();
  }
  return 0;
}

```