#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const double PI = acos(-1.0);
const int INF = 1000000000;
const int MOD = 1000000007;
class fenwick {
 public:
  vector<int> sums;
  fenwick() {}
  fenwick(int n) : sums(n, 0) {}
  int sum(int right) {
    int result = 0;
    while (right >= 0) {
      result += sums[right];
      right = (right & (right + 1)) - 1;
    }
    return result;
  }
  int sum(int from, int to) { return sum(to) - sum(from - 1); }
  void inc(int index, int delta) {
    for (; index < sums.size(); index = (index | (index + 1)))
      sums[index] += delta;
  }
};
int main() {
  int _start = clock();
  int n;
  cin >> n;
  vector<int> a(n), inv(n);
  vector<bool> bad(n, 0);
  fenwick f(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    --a[i];
    inv[a[i]] = i;
  }
  int answer = 1;
  for (int i = 1; i < n; ++i) {
    bad[i] = inv[i - 1] > inv[i];
    if (bad[i]) f.inc(i, 1);
  }
  int q, p, x, y;
  cin >> q;
  while (q--) {
    scanf("%d %d %d", &p, &x, &y);
    --x;
    --y;
    if (p == 2) {
      for (int i = max(0, a[x] - 2); i + 1 <= min(n - 1, a[x] + 2); ++i) {
        if (inv[i] > inv[i + 1]) {
          if (bad[i + 1]) {
            bad[i + 1] = 0;
            f.inc(i + 1, -1);
          }
        }
      }
      for (int i = max(0, a[y] - 2); i + 1 <= min(n - 1, a[y] + 2); ++i) {
        if (inv[i] > inv[i + 1]) {
          if (bad[i + 1]) {
            bad[i + 1] = 0;
            f.inc(i + 1, -1);
          }
        }
      }
      inv[a[x]] = y;
      inv[a[y]] = x;
      for (int i = max(0, a[x] - 2); i + 1 <= min(n - 1, a[x] + 2); ++i) {
        if (inv[i] > inv[i + 1]) {
          if (bad[i + 1] == 0) {
            bad[i + 1] = 1;
            f.inc(i + 1, +1);
          }
        }
      }
      for (int i = max(0, a[y] - 2); i + 1 <= min(n - 1, a[y] + 2); ++i) {
        if (inv[i] > inv[i + 1]) {
          if (bad[i + 1] == 0) {
            bad[i + 1] = 1;
            f.inc(i + 1, +1);
          }
        }
      }
      swap(a[x], a[y]);
      continue;
    }
    printf("%d\n", f.sum(x + 1, y) + 1);
  }
  cerr << endl
       << endl
       << "Time: " << (double)(clock() - _start) / CLOCKS_PER_SEC << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const double PI = acos(-1.0);
const int INF = 1000000000;
const int MOD = 1000000007;
class fenwick {
 public:
  vector<int> sums;
  fenwick() {}
  fenwick(int n) : sums(n, 0) {}
  int sum(int right) {
    int result = 0;
    while (right >= 0) {
      result += sums[right];
      right = (right & (right + 1)) - 1;
    }
    return result;
  }
  int sum(int from, int to) { return sum(to) - sum(from - 1); }
  void inc(int index, int delta) {
    for (; index < sums.size(); index = (index | (index + 1)))
      sums[index] += delta;
  }
};
int main() {
  int _start = clock();
  int n;
  cin >> n;
  vector<int> a(n), inv(n);
  vector<bool> bad(n, 0);
  fenwick f(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    --a[i];
    inv[a[i]] = i;
  }
  int answer = 1;
  for (int i = 1; i < n; ++i) {
    bad[i] = inv[i - 1] > inv[i];
    if (bad[i]) f.inc(i, 1);
  }
  int q, p, x, y;
  cin >> q;
  while (q--) {
    scanf("%d %d %d", &p, &x, &y);
    --x;
    --y;
    if (p == 2) {
      for (int i = max(0, a[x] - 2); i + 1 <= min(n - 1, a[x] + 2); ++i) {
        if (inv[i] > inv[i + 1]) {
          if (bad[i + 1]) {
            bad[i + 1] = 0;
            f.inc(i + 1, -1);
          }
        }
      }
      for (int i = max(0, a[y] - 2); i + 1 <= min(n - 1, a[y] + 2); ++i) {
        if (inv[i] > inv[i + 1]) {
          if (bad[i + 1]) {
            bad[i + 1] = 0;
            f.inc(i + 1, -1);
          }
        }
      }
      inv[a[x]] = y;
      inv[a[y]] = x;
      for (int i = max(0, a[x] - 2); i + 1 <= min(n - 1, a[x] + 2); ++i) {
        if (inv[i] > inv[i + 1]) {
          if (bad[i + 1] == 0) {
            bad[i + 1] = 1;
            f.inc(i + 1, +1);
          }
        }
      }
      for (int i = max(0, a[y] - 2); i + 1 <= min(n - 1, a[y] + 2); ++i) {
        if (inv[i] > inv[i + 1]) {
          if (bad[i + 1] == 0) {
            bad[i + 1] = 1;
            f.inc(i + 1, +1);
          }
        }
      }
      swap(a[x], a[y]);
      continue;
    }
    printf("%d\n", f.sum(x + 1, y) + 1);
  }
  cerr << endl
       << endl
       << "Time: " << (double)(clock() - _start) / CLOCKS_PER_SEC << endl;
  return 0;
}

```