#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 310000;
const int INF = 1 << 28;
const double EPS = 1e-9;
const double PI = 3.14159265358979;
struct BIT {
  int n;
  long long bit[MAX_N];
  void init(int n_) {
    n = n_;
    memset(bit, 0, sizeof(bit));
  }
  long long sum(int k) {
    long long s = 0;
    while (k > 0) {
      s += bit[k];
      k -= k & -k;
    }
    return s;
  }
  long long sum(int a, int b) { return sum(b) - sum(a); }
  void add(int k, long long v) {
    k++;
    while (k <= n) {
      bit[k] += v;
      k += k & -k;
    }
  }
  void set(int k, long long v) {
    add(k, -(sum(k, k + 1)));
    add(k, v);
  }
};
int n, Q;
int a[MAX_N];
int ix[MAX_N];
BIT bit;
void solve() {
  bit.init(n);
  for (int i = (0); i < (n); i++) ix[a[i]] = i;
  for (int i = (0); i < (n - 1); i++) {
    if (ix[i] > ix[i + 1]) {
      bit.set(i, 1);
    } else {
      bit.set(i, 0);
    }
  }
  for (int q = (0); q < (Q); q++) {
    int p, x, y;
    cin >> p >> x >> y;
    x--, y--;
    if (p == 1) {
      cout << bit.sum(x, y) + 1 << endl;
    } else {
      ix[a[x]] = y;
      ix[a[y]] = x;
      swap(a[x], a[y]);
      if (a[x] > 0) {
        if (ix[a[x]] < ix[a[x] - 1]) {
          bit.set(a[x] - 1, 1);
        } else {
          bit.set(a[x] - 1, 0);
        }
      }
      if (a[x] < n - 1) {
        if (ix[a[x] + 1] < ix[a[x]]) {
          bit.set(a[x], 1);
        } else {
          bit.set(a[x], 0);
        }
      }
      if (a[y] > 0) {
        if (ix[a[y]] < ix[a[y] - 1]) {
          bit.set(a[y] - 1, 1);
        } else {
          bit.set(a[y] - 1, 0);
        }
      }
      if (a[y] < n - 1) {
        if (ix[a[y] + 1] < ix[a[y]]) {
          bit.set(a[y], 1);
        } else {
          bit.set(a[y], 0);
        }
      }
    }
  }
}
int main() {
  cin >> n;
  for (int i = (0); i < (n); i++) {
    cin >> a[i];
    a[i]--;
  }
  cin >> Q;
  solve();
  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;
const int MAX_N = 310000;
const int INF = 1 << 28;
const double EPS = 1e-9;
const double PI = 3.14159265358979;
struct BIT {
  int n;
  long long bit[MAX_N];
  void init(int n_) {
    n = n_;
    memset(bit, 0, sizeof(bit));
  }
  long long sum(int k) {
    long long s = 0;
    while (k > 0) {
      s += bit[k];
      k -= k & -k;
    }
    return s;
  }
  long long sum(int a, int b) { return sum(b) - sum(a); }
  void add(int k, long long v) {
    k++;
    while (k <= n) {
      bit[k] += v;
      k += k & -k;
    }
  }
  void set(int k, long long v) {
    add(k, -(sum(k, k + 1)));
    add(k, v);
  }
};
int n, Q;
int a[MAX_N];
int ix[MAX_N];
BIT bit;
void solve() {
  bit.init(n);
  for (int i = (0); i < (n); i++) ix[a[i]] = i;
  for (int i = (0); i < (n - 1); i++) {
    if (ix[i] > ix[i + 1]) {
      bit.set(i, 1);
    } else {
      bit.set(i, 0);
    }
  }
  for (int q = (0); q < (Q); q++) {
    int p, x, y;
    cin >> p >> x >> y;
    x--, y--;
    if (p == 1) {
      cout << bit.sum(x, y) + 1 << endl;
    } else {
      ix[a[x]] = y;
      ix[a[y]] = x;
      swap(a[x], a[y]);
      if (a[x] > 0) {
        if (ix[a[x]] < ix[a[x] - 1]) {
          bit.set(a[x] - 1, 1);
        } else {
          bit.set(a[x] - 1, 0);
        }
      }
      if (a[x] < n - 1) {
        if (ix[a[x] + 1] < ix[a[x]]) {
          bit.set(a[x], 1);
        } else {
          bit.set(a[x], 0);
        }
      }
      if (a[y] > 0) {
        if (ix[a[y]] < ix[a[y] - 1]) {
          bit.set(a[y] - 1, 1);
        } else {
          bit.set(a[y] - 1, 0);
        }
      }
      if (a[y] < n - 1) {
        if (ix[a[y] + 1] < ix[a[y]]) {
          bit.set(a[y], 1);
        } else {
          bit.set(a[y], 0);
        }
      }
    }
  }
}
int main() {
  cin >> n;
  for (int i = (0); i < (n); i++) {
    cin >> a[i];
    a[i]--;
  }
  cin >> Q;
  solve();
  return 0;
}

```