#include <bits/stdc++.h>
using namespace std;
const int N = 3 * 1e5 + 100;
int n, q, a[N], sh[N], t[N], b[N];
inline int lowbit(int x) { return x & (-x); }
inline void change(int x, int v) {
  while (x <= n) sh[x] += v, x += lowbit(x);
}
inline int query(int x) {
  int ans = 0;
  while (x) ans += sh[x], x -= lowbit(x);
  return ans;
}
inline int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) b[i] = read(), a[b[i]] = i;
  for (int i = 1; i <= n; i++)
    if (a[i] < a[i - 1]) change(i, 1);
  a[n + 1] = n + 1;
  q = read();
  while (q--) {
    int op = read(), x = read(), y = read();
    if (op == 1)
      printf("%d\n", query(y) - query(x) + 1);
    else {
      int tx = x, ty = y;
      x = b[x];
      y = b[y];
      swap(b[tx], b[ty]);
      t[x] = t[x + 1] = t[y] = t[y + 1] = 0;
      if (a[x] < a[x - 1]) change(x, -1), t[x] = 1;
      if (a[x + 1] < a[x]) change(x + 1, -1), t[x + 1] = 1;
      if (a[y] < a[y - 1] && !t[y]) change(y, -1), t[y] = 1;
      if (a[y + 1] < a[y] && !t[y + 1]) change(y + 1, -1), t[y + 1] = 1;
      swap(a[x], a[y]);
      t[x] = t[x + 1] = t[y] = t[y + 1] = 0;
      if (a[x] < a[x - 1]) change(x, 1), t[x] = 1;
      if (a[x + 1] < a[x]) change(x + 1, 1), t[x + 1] = 1;
      if (a[y] < a[y - 1] && !t[y]) change(y, 1), t[y] = 1;
      if (a[y + 1] < a[y] && !t[y + 1]) change(y + 1, 1), t[y + 1] = 1;
    }
  }
}

### Prompt

Your task is to create 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 N = 3 * 1e5 + 100;
int n, q, a[N], sh[N], t[N], b[N];
inline int lowbit(int x) { return x & (-x); }
inline void change(int x, int v) {
  while (x <= n) sh[x] += v, x += lowbit(x);
}
inline int query(int x) {
  int ans = 0;
  while (x) ans += sh[x], x -= lowbit(x);
  return ans;
}
inline int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) b[i] = read(), a[b[i]] = i;
  for (int i = 1; i <= n; i++)
    if (a[i] < a[i - 1]) change(i, 1);
  a[n + 1] = n + 1;
  q = read();
  while (q--) {
    int op = read(), x = read(), y = read();
    if (op == 1)
      printf("%d\n", query(y) - query(x) + 1);
    else {
      int tx = x, ty = y;
      x = b[x];
      y = b[y];
      swap(b[tx], b[ty]);
      t[x] = t[x + 1] = t[y] = t[y + 1] = 0;
      if (a[x] < a[x - 1]) change(x, -1), t[x] = 1;
      if (a[x + 1] < a[x]) change(x + 1, -1), t[x + 1] = 1;
      if (a[y] < a[y - 1] && !t[y]) change(y, -1), t[y] = 1;
      if (a[y + 1] < a[y] && !t[y + 1]) change(y + 1, -1), t[y + 1] = 1;
      swap(a[x], a[y]);
      t[x] = t[x + 1] = t[y] = t[y + 1] = 0;
      if (a[x] < a[x - 1]) change(x, 1), t[x] = 1;
      if (a[x + 1] < a[x]) change(x + 1, 1), t[x + 1] = 1;
      if (a[y] < a[y - 1] && !t[y]) change(y, 1), t[y] = 1;
      if (a[y + 1] < a[y] && !t[y + 1]) change(y + 1, 1), t[y + 1] = 1;
    }
  }
}

```