#include <bits/stdc++.h>
using namespace std;
template <class T>
void setmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
void setmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
const int MAXN = 300010;
int tr[MAXN], t[MAXN];
int pos[MAXN], a[MAXN];
int n, qs;
void add(int x, int v) {
  for (; x <= n; x += (x & -x)) tr[x] += v;
}
int get(int x) {
  int ans = 0;
  for (; x >= 1; x -= (x & -x)) ans += tr[x];
  return ans;
}
void upt(int x) {
  if (x <= 1 || x > n) return;
  add(x, -t[x]);
  if (pos[x] < pos[x - 1])
    t[x] = 1;
  else
    t[x] = 0;
  add(x, t[x]);
}
int main() {
  scanf("%d", &n);
  for (int i = (int)(1); i <= (int)(n); ++i) scanf("%d", &a[i]), pos[a[i]] = i;
  memset(tr, 0, sizeof(tr));
  t[1] = 0;
  for (int i = (int)(2); i <= (int)(n); ++i) {
    if (pos[i] < pos[i - 1])
      t[i] = 1;
    else
      t[i] = 0;
    add(i, t[i]);
  }
  scanf("%d", &qs);
  while (qs--) {
    int p, x, y;
    scanf("%d%d%d", &p, &x, &y);
    if (p == 1) {
      int ans = get(y) - get(x) + 1;
      printf("%d\n", ans);
    } else {
      swap(a[x], a[y]);
      pos[a[x]] = x;
      pos[a[y]] = y;
      x = a[x], y = a[y];
      upt(x);
      upt(x + 1);
      upt(y);
      upt(y + 1);
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
template <class T>
void setmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
void setmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
const int MAXN = 300010;
int tr[MAXN], t[MAXN];
int pos[MAXN], a[MAXN];
int n, qs;
void add(int x, int v) {
  for (; x <= n; x += (x & -x)) tr[x] += v;
}
int get(int x) {
  int ans = 0;
  for (; x >= 1; x -= (x & -x)) ans += tr[x];
  return ans;
}
void upt(int x) {
  if (x <= 1 || x > n) return;
  add(x, -t[x]);
  if (pos[x] < pos[x - 1])
    t[x] = 1;
  else
    t[x] = 0;
  add(x, t[x]);
}
int main() {
  scanf("%d", &n);
  for (int i = (int)(1); i <= (int)(n); ++i) scanf("%d", &a[i]), pos[a[i]] = i;
  memset(tr, 0, sizeof(tr));
  t[1] = 0;
  for (int i = (int)(2); i <= (int)(n); ++i) {
    if (pos[i] < pos[i - 1])
      t[i] = 1;
    else
      t[i] = 0;
    add(i, t[i]);
  }
  scanf("%d", &qs);
  while (qs--) {
    int p, x, y;
    scanf("%d%d%d", &p, &x, &y);
    if (p == 1) {
      int ans = get(y) - get(x) + 1;
      printf("%d\n", ans);
    } else {
      swap(a[x], a[y]);
      pos[a[x]] = x;
      pos[a[y]] = y;
      x = a[x], y = a[y];
      upt(x);
      upt(x + 1);
      upt(y);
      upt(y + 1);
    }
  }
  return 0;
}

```