#include <bits/stdc++.h>
using namespace std;
const int NMax = 301000, MMax = 200;
struct node {
  int size;
  int leftmost, rightmost;
} T[NMax * 4];
int N;
void insert(int a, int b, int first, int second, int p) {
  if (first == second) {
    T[p].leftmost = T[p].rightmost = b;
    return;
  }
  int mid = (first + second) >> 1;
  if (mid >= a)
    insert(a, b, first, mid, p * 2);
  else
    insert(a, b, mid + 1, second, p * 2 + 1);
  T[p].leftmost = T[p * 2].leftmost;
  T[p].rightmost = T[p * 2 + 1].rightmost;
  T[p].size = T[p * 2].size + T[p * 2 + 1].size;
  if (T[p * 2].rightmost > T[p * 2 + 1].leftmost) T[p].size++;
}
int calc(int a, int b, int first, int second, int p) {
  if (first == a && second == b) return T[p].size;
  int mid = (first + second) >> 1;
  if (mid >= b)
    return calc(a, b, first, mid, p * 2);
  else if (mid < a)
    return calc(a, b, mid + 1, second, p * 2 + 1);
  else {
    int ret = calc(a, mid, first, mid, p * 2) +
              calc(mid + 1, b, mid + 1, second, p * 2 + 1);
    if (T[p * 2].rightmost > T[p * 2 + 1].leftmost) ret++;
    return ret;
  }
}
int A[NMax];
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    int first;
    scanf("%d", &first);
    insert(first, i, 1, N, 1);
    A[i] = first;
  }
  int Q;
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) {
    int t, first, second;
    scanf("%d%d%d", &t, &first, &second);
    if (t == 1)
      printf("%d\n", calc(first, second, 1, N, 1) + 1);
    else {
      swap(A[first], A[second]);
      insert(A[first], first, 1, N, 1);
      insert(A[second], second, 1, N, 1);
    }
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 NMax = 301000, MMax = 200;
struct node {
  int size;
  int leftmost, rightmost;
} T[NMax * 4];
int N;
void insert(int a, int b, int first, int second, int p) {
  if (first == second) {
    T[p].leftmost = T[p].rightmost = b;
    return;
  }
  int mid = (first + second) >> 1;
  if (mid >= a)
    insert(a, b, first, mid, p * 2);
  else
    insert(a, b, mid + 1, second, p * 2 + 1);
  T[p].leftmost = T[p * 2].leftmost;
  T[p].rightmost = T[p * 2 + 1].rightmost;
  T[p].size = T[p * 2].size + T[p * 2 + 1].size;
  if (T[p * 2].rightmost > T[p * 2 + 1].leftmost) T[p].size++;
}
int calc(int a, int b, int first, int second, int p) {
  if (first == a && second == b) return T[p].size;
  int mid = (first + second) >> 1;
  if (mid >= b)
    return calc(a, b, first, mid, p * 2);
  else if (mid < a)
    return calc(a, b, mid + 1, second, p * 2 + 1);
  else {
    int ret = calc(a, mid, first, mid, p * 2) +
              calc(mid + 1, b, mid + 1, second, p * 2 + 1);
    if (T[p * 2].rightmost > T[p * 2 + 1].leftmost) ret++;
    return ret;
  }
}
int A[NMax];
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    int first;
    scanf("%d", &first);
    insert(first, i, 1, N, 1);
    A[i] = first;
  }
  int Q;
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) {
    int t, first, second;
    scanf("%d%d%d", &t, &first, &second);
    if (t == 1)
      printf("%d\n", calc(first, second, 1, N, 1) + 1);
    else {
      swap(A[first], A[second]);
      insert(A[first], first, 1, N, 1);
      insert(A[second], second, 1, N, 1);
    }
  }
  return 0;
}

```