#include <bits/stdc++.h>
using namespace std;
int N[300005], P[300005];
int T[300005];
void Add(int i, int m) {
  for (; i < 300005; i += i & (-i)) T[i] += m;
}
int Sum(int i) {
  int t, sum = 0;
  for (; i; i -= i & (-i)) sum += T[i];
  return sum;
}
int Count_sum(int i, int j) { return Sum(j) - Sum(i - 1); }
int Work(int x, int y) {
  int res = 0, i, bef = 300005;
  res = Count_sum(x + 1, y);
  return res;
}
void Done(int x, int y) {
  if (P[x - 1] < P[x] && P[x - 1] >= P[y]) Add(x, 1);
  if (P[x - 1] > P[x] && P[x - 1] <= P[y]) Add(x, -1);
  if (P[x] > P[x + 1] && P[y] <= P[x + 1] && y != x + 1) Add(x + 1, -1);
  if (P[x] < P[x + 1] && P[y] >= P[x + 1] && y != x + 1) Add(x + 1, 1);
}
void Do(int x, int y) {
  Done(x, y);
  Done(y, x);
}
int main() {
  int i, j, n, m;
  for (i = scanf("%d", &n); i <= n; i++) scanf("%d", &N[i]), P[N[i]] = i;
  for (i = 1; i <= n; i++)
    if (P[i] < P[i - 1]) Add(i, 1);
  int p, x, y;
  for (i = scanf("%d", &n); i <= n; i++) {
    scanf("%d%d%d", &p, &x, &y);
    if (p == 1) {
      printf("%d\n", Work(x, y) + 1);
    } else
      Do(N[x], N[y]), swap(N[x], N[y]), P[N[x]] = x, P[N[y]] = y;
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
int N[300005], P[300005];
int T[300005];
void Add(int i, int m) {
  for (; i < 300005; i += i & (-i)) T[i] += m;
}
int Sum(int i) {
  int t, sum = 0;
  for (; i; i -= i & (-i)) sum += T[i];
  return sum;
}
int Count_sum(int i, int j) { return Sum(j) - Sum(i - 1); }
int Work(int x, int y) {
  int res = 0, i, bef = 300005;
  res = Count_sum(x + 1, y);
  return res;
}
void Done(int x, int y) {
  if (P[x - 1] < P[x] && P[x - 1] >= P[y]) Add(x, 1);
  if (P[x - 1] > P[x] && P[x - 1] <= P[y]) Add(x, -1);
  if (P[x] > P[x + 1] && P[y] <= P[x + 1] && y != x + 1) Add(x + 1, -1);
  if (P[x] < P[x + 1] && P[y] >= P[x + 1] && y != x + 1) Add(x + 1, 1);
}
void Do(int x, int y) {
  Done(x, y);
  Done(y, x);
}
int main() {
  int i, j, n, m;
  for (i = scanf("%d", &n); i <= n; i++) scanf("%d", &N[i]), P[N[i]] = i;
  for (i = 1; i <= n; i++)
    if (P[i] < P[i - 1]) Add(i, 1);
  int p, x, y;
  for (i = scanf("%d", &n); i <= n; i++) {
    scanf("%d%d%d", &p, &x, &y);
    if (p == 1) {
      printf("%d\n", Work(x, y) + 1);
    } else
      Do(N[x], N[y]), swap(N[x], N[y]), P[N[x]] = x, P[N[y]] = y;
  }
  return 0;
}

```