#include <bits/stdc++.h>
using namespace std;
map<long long, int> ilk, son;
vector<int> V;
int N;
int A[300005], B[300005], ST[4 * 300005];
int init(int pos, int s, int e) {
  if (s == e) return ST[pos] = (B[s] < B[s - 1]);
  int m = (s + e) >> 1;
  return ST[pos] = init(pos * 2, s, m) + init(pos * 2 + 1, m + 1, e);
}
int find(int pos, int s, int e, int a, int b) {
  if (a > e || b < s) return 0;
  if (a <= s && e <= b) return ST[pos];
  int m = (s + e) >> 1;
  return find(pos * 2, s, m, a, b) + find(pos * 2 + 1, m + 1, e, a, b);
}
int update(int pos, int s, int e, int a, int v) {
  if (a < s || a > e) return ST[pos];
  if (s == e) return ST[pos] = v;
  int m = (s + e) >> 1;
  return ST[pos] =
             update(pos * 2, s, m, a, v) + update(pos * 2 + 1, m + 1, e, a, v);
}
void solve() {
  int Q, a, b, i, k, k1, k2;
  scanf("%d", &N);
  for (i = 1; i <= N; i++) {
    scanf("%d", A + i);
    B[A[i]] = i;
  }
  init(1, 1, N);
  scanf("%d", &Q);
  B[N + 1] = N + 100;
  for (i = 1; i <= Q; i++) {
    scanf("%d %d %d", &k, &a, &b);
    if (a > b) swap(a, b);
    if (k == 1) {
      if (a == b)
        printf("1\n");
      else
        printf("%d\n", find(1, 1, N, a + 1, b) + 1);
    } else {
      k1 = A[a];
      k2 = A[b];
      if (k1 + 1 == k2) {
        update(1, 1, N, k1, (b < B[k1 - 1]));
        update(1, 1, N, k2, (b > a));
        update(1, 1, N, k2 + 1, (a > B[k2 + 1]));
      } else {
        update(1, 1, N, k1, (b < B[k1 - 1]));
        update(1, 1, N, k1 + 1, (b > B[k1 + 1]));
        update(1, 1, N, k2, (a < B[k2 - 1]));
        update(1, 1, N, k2 + 1, (a > B[k2 + 1]));
      }
      swap(A[a], A[b]);
      swap(B[k1], B[k2]);
    }
  }
}
int main() {
  solve();
  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;
map<long long, int> ilk, son;
vector<int> V;
int N;
int A[300005], B[300005], ST[4 * 300005];
int init(int pos, int s, int e) {
  if (s == e) return ST[pos] = (B[s] < B[s - 1]);
  int m = (s + e) >> 1;
  return ST[pos] = init(pos * 2, s, m) + init(pos * 2 + 1, m + 1, e);
}
int find(int pos, int s, int e, int a, int b) {
  if (a > e || b < s) return 0;
  if (a <= s && e <= b) return ST[pos];
  int m = (s + e) >> 1;
  return find(pos * 2, s, m, a, b) + find(pos * 2 + 1, m + 1, e, a, b);
}
int update(int pos, int s, int e, int a, int v) {
  if (a < s || a > e) return ST[pos];
  if (s == e) return ST[pos] = v;
  int m = (s + e) >> 1;
  return ST[pos] =
             update(pos * 2, s, m, a, v) + update(pos * 2 + 1, m + 1, e, a, v);
}
void solve() {
  int Q, a, b, i, k, k1, k2;
  scanf("%d", &N);
  for (i = 1; i <= N; i++) {
    scanf("%d", A + i);
    B[A[i]] = i;
  }
  init(1, 1, N);
  scanf("%d", &Q);
  B[N + 1] = N + 100;
  for (i = 1; i <= Q; i++) {
    scanf("%d %d %d", &k, &a, &b);
    if (a > b) swap(a, b);
    if (k == 1) {
      if (a == b)
        printf("1\n");
      else
        printf("%d\n", find(1, 1, N, a + 1, b) + 1);
    } else {
      k1 = A[a];
      k2 = A[b];
      if (k1 + 1 == k2) {
        update(1, 1, N, k1, (b < B[k1 - 1]));
        update(1, 1, N, k2, (b > a));
        update(1, 1, N, k2 + 1, (a > B[k2 + 1]));
      } else {
        update(1, 1, N, k1, (b < B[k1 - 1]));
        update(1, 1, N, k1 + 1, (b > B[k1 + 1]));
        update(1, 1, N, k2, (a < B[k2 - 1]));
        update(1, 1, N, k2 + 1, (a > B[k2 + 1]));
      }
      swap(A[a], A[b]);
      swap(B[k1], B[k2]);
    }
  }
}
int main() {
  solve();
  return 0;
}

```