#include <bits/stdc++.h>
using namespace std;
template <typename T>
T mabs(const T &a) {
  return a < 0 ? -a : a;
}
const int SZ = 300300;
int End[SZ], Beg[SZ];
void addFen(int pos, int val, int *Fen) {
  for (int i = pos; i < SZ; i += i & -i) Fen[i] += val;
}
int getFen(int pos, int *Fen) {
  int res = 0;
  for (int i = pos; i > 0; i -= i & -i) res += Fen[i];
  return res;
}
int getAB(int a, int b, int *Fen) {
  int res = getFen(b, Fen);
  if (a > 1) res -= getFen(a - 1, Fen);
  return res;
}
map<int, int> mp;
int n;
int A[SZ];
void check(int i) {
  if (i == n || mp[i + 1] < mp[i]) addFen(i, 1, End);
  if (i == 1 || mp[i - 1] > mp[i]) addFen(i, 1, Beg);
}
void uncheck(int i) {
  if (getAB(i, i, Beg)) addFen(i, -1, Beg);
  if (getAB(i, i, End)) addFen(i, -1, End);
}
void proc(int a, int b) {
  int valB = A[b];
  int valA = A[a];
  swap(A[a], A[b]);
  mp[valA] = b;
  mp[valB] = a;
  vector<int> v;
  v.push_back(valA);
  v.push_back(valB);
  for (int i = (0), ei = (2); i < ei; i++)
    for (int j = -1; j <= 1; j += 2)
      if (v[i] + j >= 1 && v[i] + j <= n) v.push_back(v[i] + j);
  sort((v).begin(), (v).end());
  v.resize(unique((v).begin(), (v).end()) - v.begin());
  for (int i = (0), ei = (v.size()); i < ei; i++) uncheck(v[i]);
  for (int i = (0), ei = (v.size()); i < ei; i++) check(v[i]);
}
void show(int n) {
  puts("");
  for (int i = (1), ei = (n + 1); i < ei; i++) printf("%d ", getAB(i, i, Beg));
  puts("");
  for (int i = (1), ei = (n + 1); i < ei; i++) printf("%d ", getAB(i, i, End));
}
void run() {
  cin >> n;
  for (int i = (0), ei = (n); i < ei; i++) {
    scanf("%d", A + i);
    mp[A[i]] = i;
  }
  for (int i = (1), ei = (n + 1); i < ei; i++) {
    check(i);
  }
  int q;
  cin >> q;
  for (int i = (0), ei = (q); i < ei; i++) {
    int t, a, b;
    scanf("%d%d%d", &t, &a, &b);
    if (t == 1) {
      int res = getAB(a, b, End);
      if (a != n && mp[b + 1] > mp[b]) res++;
      printf("%d\n", res);
    } else {
      --a;
      --b;
      proc(a, b);
    }
  }
}
int main() {
  run();
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 <typename T>
T mabs(const T &a) {
  return a < 0 ? -a : a;
}
const int SZ = 300300;
int End[SZ], Beg[SZ];
void addFen(int pos, int val, int *Fen) {
  for (int i = pos; i < SZ; i += i & -i) Fen[i] += val;
}
int getFen(int pos, int *Fen) {
  int res = 0;
  for (int i = pos; i > 0; i -= i & -i) res += Fen[i];
  return res;
}
int getAB(int a, int b, int *Fen) {
  int res = getFen(b, Fen);
  if (a > 1) res -= getFen(a - 1, Fen);
  return res;
}
map<int, int> mp;
int n;
int A[SZ];
void check(int i) {
  if (i == n || mp[i + 1] < mp[i]) addFen(i, 1, End);
  if (i == 1 || mp[i - 1] > mp[i]) addFen(i, 1, Beg);
}
void uncheck(int i) {
  if (getAB(i, i, Beg)) addFen(i, -1, Beg);
  if (getAB(i, i, End)) addFen(i, -1, End);
}
void proc(int a, int b) {
  int valB = A[b];
  int valA = A[a];
  swap(A[a], A[b]);
  mp[valA] = b;
  mp[valB] = a;
  vector<int> v;
  v.push_back(valA);
  v.push_back(valB);
  for (int i = (0), ei = (2); i < ei; i++)
    for (int j = -1; j <= 1; j += 2)
      if (v[i] + j >= 1 && v[i] + j <= n) v.push_back(v[i] + j);
  sort((v).begin(), (v).end());
  v.resize(unique((v).begin(), (v).end()) - v.begin());
  for (int i = (0), ei = (v.size()); i < ei; i++) uncheck(v[i]);
  for (int i = (0), ei = (v.size()); i < ei; i++) check(v[i]);
}
void show(int n) {
  puts("");
  for (int i = (1), ei = (n + 1); i < ei; i++) printf("%d ", getAB(i, i, Beg));
  puts("");
  for (int i = (1), ei = (n + 1); i < ei; i++) printf("%d ", getAB(i, i, End));
}
void run() {
  cin >> n;
  for (int i = (0), ei = (n); i < ei; i++) {
    scanf("%d", A + i);
    mp[A[i]] = i;
  }
  for (int i = (1), ei = (n + 1); i < ei; i++) {
    check(i);
  }
  int q;
  cin >> q;
  for (int i = (0), ei = (q); i < ei; i++) {
    int t, a, b;
    scanf("%d%d%d", &t, &a, &b);
    if (t == 1) {
      int res = getAB(a, b, End);
      if (a != n && mp[b + 1] > mp[b]) res++;
      printf("%d\n", res);
    } else {
      --a;
      --b;
      proc(a, b);
    }
  }
}
int main() {
  run();
  return 0;
}

```