#include <bits/stdc++.h>
using namespace std;
const int MAXN = 300005;
class BIT {
  int range;
  int *entries;

 public:
  BIT(const int &_range) : range(_range) {
    entries = new int[range + 1];
    fill(entries, entries + range + 1, 0);
  }
  ~BIT() { delete[] entries; }
  void insert(int x, int val) {
    for (int i = x; i <= range; i += i & (-i)) {
      entries[i] += val;
    }
  }
  int query(int x) {
    int res = 0;
    for (int i = x; i >= 1; i -= i & (-i)) {
      res += entries[i];
    }
    return res;
  }
  int query(int st, int ed) {
    assert(st <= ed);
    st--;
    int res_st = (st >= 1) ? query(st) : 0;
    int res_ed = (ed >= 1) ? query(ed) : 0;
    return res_ed - res_st;
  }
};
int n;
int q;
int a[MAXN];
int pos[MAXN];
int pre[MAXN];
int nxt[MAXN];
int app[MAXN];
set<int> curr_set;
BIT bit(MAXN);
void update_sub(int x) {
  if (pre[x] != -1) {
    if (pos[x] > pos[x - 1]) {
      nxt[x - 1] = pos[x];
    } else {
      nxt[x - 1] = -1;
      pre[x] = -1;
      if (curr_set.find(x) == curr_set.end()) {
        curr_set.insert(x);
        bit.insert(x, 1);
      }
    }
  }
  if (nxt[x] != -1) {
    if (pos[x + 1] > pos[x]) {
      pre[x + 1] = pos[x];
    } else {
      nxt[x] = -1;
      pre[x + 1] = -1;
      if (curr_set.find(x + 1) == curr_set.end()) {
        curr_set.insert(x + 1);
        bit.insert(x + 1, 1);
      }
    }
  }
}
void merge(int x) {
  if (pre[x] != -1 && nxt[x] != -1) return;
  if (pre[x] == -1 && x > 1) {
    if (pos[x - 1] < pos[x]) {
      pre[x] = pos[x - 1];
      nxt[x - 1] = pos[x];
      curr_set.erase(x);
      bit.insert(x, -1);
    }
  }
  if (nxt[x] == -1 && x < n) {
    if (pos[x + 1] > pos[x]) {
      pre[x + 1] = pos[x];
      nxt[x] = pos[x + 1];
      curr_set.erase(x + 1);
      bit.insert(x + 1, -1);
    }
  }
}
void update(int i, int j) {
  int x = a[i];
  int y = a[j];
  pos[x] = j;
  pos[y] = i;
  swap(a[i], a[j]);
  update_sub(x);
  update_sub(y);
  merge(x);
  merge(y);
}
int main() {
  scanf("%d", &(n));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &(a[i]));
    pos[a[i]] = i;
  }
  memset((pre), -1, sizeof((pre)));
  memset((nxt), -1, sizeof((nxt)));
  memset((app), -1, sizeof((app)));
  for (int i = 1; i <= n; i++) {
    if (app[a[i]] == -1) {
      app[a[i]] = 1;
      app[a[i] + 1] = 1;
      bit.insert(a[i], 1);
    } else {
      nxt[a[i] - 1] = i;
      pre[a[i]] = pos[a[i] - 1];
      app[a[i] + 1] = 1;
    }
  }
  scanf("%d", &(q));
  while (q-- > 0) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int st, ed;
      scanf("%d%d", &st, &ed);
      int ans = bit.query(st, ed);
      if (bit.query(st, st) == 0) ans++;
      printf("%d\n", ans);
    } else {
      int i, j;
      scanf("%d%d", &i, &j);
      if (i == j) continue;
      update(i, j);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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 MAXN = 300005;
class BIT {
  int range;
  int *entries;

 public:
  BIT(const int &_range) : range(_range) {
    entries = new int[range + 1];
    fill(entries, entries + range + 1, 0);
  }
  ~BIT() { delete[] entries; }
  void insert(int x, int val) {
    for (int i = x; i <= range; i += i & (-i)) {
      entries[i] += val;
    }
  }
  int query(int x) {
    int res = 0;
    for (int i = x; i >= 1; i -= i & (-i)) {
      res += entries[i];
    }
    return res;
  }
  int query(int st, int ed) {
    assert(st <= ed);
    st--;
    int res_st = (st >= 1) ? query(st) : 0;
    int res_ed = (ed >= 1) ? query(ed) : 0;
    return res_ed - res_st;
  }
};
int n;
int q;
int a[MAXN];
int pos[MAXN];
int pre[MAXN];
int nxt[MAXN];
int app[MAXN];
set<int> curr_set;
BIT bit(MAXN);
void update_sub(int x) {
  if (pre[x] != -1) {
    if (pos[x] > pos[x - 1]) {
      nxt[x - 1] = pos[x];
    } else {
      nxt[x - 1] = -1;
      pre[x] = -1;
      if (curr_set.find(x) == curr_set.end()) {
        curr_set.insert(x);
        bit.insert(x, 1);
      }
    }
  }
  if (nxt[x] != -1) {
    if (pos[x + 1] > pos[x]) {
      pre[x + 1] = pos[x];
    } else {
      nxt[x] = -1;
      pre[x + 1] = -1;
      if (curr_set.find(x + 1) == curr_set.end()) {
        curr_set.insert(x + 1);
        bit.insert(x + 1, 1);
      }
    }
  }
}
void merge(int x) {
  if (pre[x] != -1 && nxt[x] != -1) return;
  if (pre[x] == -1 && x > 1) {
    if (pos[x - 1] < pos[x]) {
      pre[x] = pos[x - 1];
      nxt[x - 1] = pos[x];
      curr_set.erase(x);
      bit.insert(x, -1);
    }
  }
  if (nxt[x] == -1 && x < n) {
    if (pos[x + 1] > pos[x]) {
      pre[x + 1] = pos[x];
      nxt[x] = pos[x + 1];
      curr_set.erase(x + 1);
      bit.insert(x + 1, -1);
    }
  }
}
void update(int i, int j) {
  int x = a[i];
  int y = a[j];
  pos[x] = j;
  pos[y] = i;
  swap(a[i], a[j]);
  update_sub(x);
  update_sub(y);
  merge(x);
  merge(y);
}
int main() {
  scanf("%d", &(n));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &(a[i]));
    pos[a[i]] = i;
  }
  memset((pre), -1, sizeof((pre)));
  memset((nxt), -1, sizeof((nxt)));
  memset((app), -1, sizeof((app)));
  for (int i = 1; i <= n; i++) {
    if (app[a[i]] == -1) {
      app[a[i]] = 1;
      app[a[i] + 1] = 1;
      bit.insert(a[i], 1);
    } else {
      nxt[a[i] - 1] = i;
      pre[a[i]] = pos[a[i] - 1];
      app[a[i] + 1] = 1;
    }
  }
  scanf("%d", &(q));
  while (q-- > 0) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int st, ed;
      scanf("%d%d", &st, &ed);
      int ans = bit.query(st, ed);
      if (bit.query(st, st) == 0) ans++;
      printf("%d\n", ans);
    } else {
      int i, j;
      scanf("%d%d", &i, &j);
      if (i == j) continue;
      update(i, j);
    }
  }
  return 0;
}

```