#include <bits/stdc++.h>
using namespace std;
template <class T>
T gcd(T x, T y) {
  while (T t = x % y) x = y, y = t;
  return y;
}
const double eps = 1e-10;
const double PI = acos(-1.);
const int INF = 1000000000;
const int MOD = 1000000007;
const double E = 2.7182818284590452353602874713527;
bool isdig(char x) { return x >= '0' && x <= '9'; }
bool isup(char x) { return x >= 'A' && x <= 'Z'; }
bool islow(char x) { return x >= 'a' && x <= 'z'; }
bool islet(char x) { return isup(x) || islow(x); }
int N, num[100005];
int loc[100005], bef[100005];
void insert(int x, int add) {
  while (x <= N) {
    bef[x] += add;
    x += ((x) & (-(x)));
  }
}
void get_data() {
  scanf("%d", &N);
  memset(num, -1, sizeof(num));
  ;
  for (int i = 0; i < N; i++) scanf("%d", &num[i + 1]);
  memset(loc, -1, sizeof(loc));
  ;
  for (int i = 1; i <= N; i++) loc[num[i]] = i;
  memset(bef, 0, sizeof(bef));
  ;
  for (int i = 2; i <= N; i++)
    if (loc[i] < loc[i - 1]) insert(i, 1);
}
int qry(int x) {
  int r = 0;
  while (x) {
    r += bef[x];
    x -= ((x) & (-(x)));
  }
  return r;
}
int qry(int s, int e) { return qry(e) - qry(s); }
void run() {
  int q, a, b;
  cin >> q;
  while (q--) {
    int t;
    scanf("%d%d%d", &t, &a, &b);
    if (t == 1) {
      printf("%d\n", 1 + qry(a, b));
    } else {
      vector<int> vec;
      vec.push_back(num[a]);
      vec.push_back(num[b]);
      int ini[4], now[4];
      if (num[a] + 1 != num[b]) vec.push_back(num[a] + 1);
      if (num[b] + 1 != num[a]) vec.push_back(num[b] + 1);
      for (int i = 0; i < ((int)(vec).size()); i++) {
        ini[i] = vec[i] > 1 && loc[vec[i]] < loc[vec[i] - 1];
      }
      loc[num[a]] = b;
      loc[num[b]] = a;
      swap(num[a], num[b]);
      for (int i = 0; i < ((int)(vec).size()); i++) {
        now[i] = vec[i] > 1 && loc[vec[i]] < loc[vec[i] - 1];
        insert(vec[i], now[i] - ini[i]);
      }
    }
  }
}
int main(int argc, char *argv[]) {
  {
    get_data();
    run();
  }
  return EXIT_SUCCESS;
}

### Prompt

Develop a solution in Cpp 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 <class T>
T gcd(T x, T y) {
  while (T t = x % y) x = y, y = t;
  return y;
}
const double eps = 1e-10;
const double PI = acos(-1.);
const int INF = 1000000000;
const int MOD = 1000000007;
const double E = 2.7182818284590452353602874713527;
bool isdig(char x) { return x >= '0' && x <= '9'; }
bool isup(char x) { return x >= 'A' && x <= 'Z'; }
bool islow(char x) { return x >= 'a' && x <= 'z'; }
bool islet(char x) { return isup(x) || islow(x); }
int N, num[100005];
int loc[100005], bef[100005];
void insert(int x, int add) {
  while (x <= N) {
    bef[x] += add;
    x += ((x) & (-(x)));
  }
}
void get_data() {
  scanf("%d", &N);
  memset(num, -1, sizeof(num));
  ;
  for (int i = 0; i < N; i++) scanf("%d", &num[i + 1]);
  memset(loc, -1, sizeof(loc));
  ;
  for (int i = 1; i <= N; i++) loc[num[i]] = i;
  memset(bef, 0, sizeof(bef));
  ;
  for (int i = 2; i <= N; i++)
    if (loc[i] < loc[i - 1]) insert(i, 1);
}
int qry(int x) {
  int r = 0;
  while (x) {
    r += bef[x];
    x -= ((x) & (-(x)));
  }
  return r;
}
int qry(int s, int e) { return qry(e) - qry(s); }
void run() {
  int q, a, b;
  cin >> q;
  while (q--) {
    int t;
    scanf("%d%d%d", &t, &a, &b);
    if (t == 1) {
      printf("%d\n", 1 + qry(a, b));
    } else {
      vector<int> vec;
      vec.push_back(num[a]);
      vec.push_back(num[b]);
      int ini[4], now[4];
      if (num[a] + 1 != num[b]) vec.push_back(num[a] + 1);
      if (num[b] + 1 != num[a]) vec.push_back(num[b] + 1);
      for (int i = 0; i < ((int)(vec).size()); i++) {
        ini[i] = vec[i] > 1 && loc[vec[i]] < loc[vec[i] - 1];
      }
      loc[num[a]] = b;
      loc[num[b]] = a;
      swap(num[a], num[b]);
      for (int i = 0; i < ((int)(vec).size()); i++) {
        now[i] = vec[i] > 1 && loc[vec[i]] < loc[vec[i] - 1];
        insert(vec[i], now[i] - ini[i]);
      }
    }
  }
}
int main(int argc, char *argv[]) {
  {
    get_data();
    run();
  }
  return EXIT_SUCCESS;
}

```