#include <bits/stdc++.h>
using namespace std;
struct BIT {
  int dat[300010];
  void ins(int x, int v) {
    for (int i = x; i < 300010; i += i & -i) dat[i] += v;
  }
  int ask(int x) {
    int s = 0;
    for (int i = x; i; i -= i & -i) s += dat[i];
    return s;
  }
} bit;
int n, m, a[300010], p[300010];
set<pair<int, int> > s;
void s_ins(int l, int r) {
  bit.ins(l, +1);
  s.insert(pair<int, int>(l, r));
}
void s_del(set<pair<int, int> >::const_iterator it) {
  bit.ins(it->first, -1);
  s.erase(it);
}
void del(int x) {
  auto it = s.upper_bound(pair<int, int>(x, 300010));
  it--;
  int l = it->first;
  int r = it->second;
  s_del(it);
  if (l <= x - 1) s_ins(l, x - 1);
  s_ins(x, x);
  if (x + 1 <= r) s_ins(x + 1, r);
}
void ins(int x) {
  {
    auto it = s.upper_bound(pair<int, int>(x, 300010));
    it--;
    if (it != s.begin()) {
      auto jt = it;
      jt--;
      if (p[jt->second] < p[it->first]) {
        int l = jt->first;
        int r = it->second;
        s_del(it);
        s_del(jt);
        s_ins(l, r);
      }
    }
  }
  {
    auto it = s.upper_bound(pair<int, int>(x, 300010));
    if (it != s.end()) {
      auto jt = it;
      jt--;
      if (p[jt->second] < p[it->first]) {
        int l = jt->first;
        int r = it->second;
        s_del(it);
        s_del(jt);
        s_ins(l, r);
      }
    }
  }
}
int get_left(int x) {
  auto it = s.upper_bound(pair<int, int>(x, 300010));
  it--;
  return it->first;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++) p[a[i]] = i;
  for (int i = 1, j; i <= n; i = j) {
    for (j = i + 1; j <= n && p[j] >= p[j - 1]; j++)
      ;
    s_ins(i, j - 1);
  }
  scanf("%d", &m);
  while (m--) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      int l = get_left(x);
      int r = get_left(y);
      printf("%d\n", bit.ask(r) - bit.ask(l - 1));
    } else {
      del(a[x]);
      del(a[y]);
      swap(a[x], a[y]);
      swap(p[a[x]], p[a[y]]);
      ins(a[x]);
      ins(a[y]);
    }
  }
  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;
struct BIT {
  int dat[300010];
  void ins(int x, int v) {
    for (int i = x; i < 300010; i += i & -i) dat[i] += v;
  }
  int ask(int x) {
    int s = 0;
    for (int i = x; i; i -= i & -i) s += dat[i];
    return s;
  }
} bit;
int n, m, a[300010], p[300010];
set<pair<int, int> > s;
void s_ins(int l, int r) {
  bit.ins(l, +1);
  s.insert(pair<int, int>(l, r));
}
void s_del(set<pair<int, int> >::const_iterator it) {
  bit.ins(it->first, -1);
  s.erase(it);
}
void del(int x) {
  auto it = s.upper_bound(pair<int, int>(x, 300010));
  it--;
  int l = it->first;
  int r = it->second;
  s_del(it);
  if (l <= x - 1) s_ins(l, x - 1);
  s_ins(x, x);
  if (x + 1 <= r) s_ins(x + 1, r);
}
void ins(int x) {
  {
    auto it = s.upper_bound(pair<int, int>(x, 300010));
    it--;
    if (it != s.begin()) {
      auto jt = it;
      jt--;
      if (p[jt->second] < p[it->first]) {
        int l = jt->first;
        int r = it->second;
        s_del(it);
        s_del(jt);
        s_ins(l, r);
      }
    }
  }
  {
    auto it = s.upper_bound(pair<int, int>(x, 300010));
    if (it != s.end()) {
      auto jt = it;
      jt--;
      if (p[jt->second] < p[it->first]) {
        int l = jt->first;
        int r = it->second;
        s_del(it);
        s_del(jt);
        s_ins(l, r);
      }
    }
  }
}
int get_left(int x) {
  auto it = s.upper_bound(pair<int, int>(x, 300010));
  it--;
  return it->first;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++) p[a[i]] = i;
  for (int i = 1, j; i <= n; i = j) {
    for (j = i + 1; j <= n && p[j] >= p[j - 1]; j++)
      ;
    s_ins(i, j - 1);
  }
  scanf("%d", &m);
  while (m--) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      int l = get_left(x);
      int r = get_left(y);
      printf("%d\n", bit.ask(r) - bit.ask(l - 1));
    } else {
      del(a[x]);
      del(a[y]);
      swap(a[x], a[y]);
      swap(p[a[x]], p[a[y]]);
      ins(a[x]);
      ins(a[y]);
    }
  }
  return 0;
}

```