#include <bits/stdc++.h>
using namespace std;
int n, q, op;
set<pair<int, int> > s;
int x, y;
int bit[300010];
int v[300010];
void update(int x, int v) {
  while (x < 300010) {
    bit[x] += v;
    x += (x & -x);
  }
}
int query(int x) {
  int sum = 0;
  while (x > 0) {
    sum += bit[x];
    x -= (x & -x);
  }
  return sum;
}
void printset() {
  printf("Set -------- \n");
  for (auto it = s.begin(); it != s.end(); ++it) {
    printf("%d - %d\n", it->first, it->second);
  }
}
int main(void) {
  ios ::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> v[i];
    s.insert(make_pair(v[i], i));
  }
  for (auto it = s.begin(); it != s.end(); ++it) {
    if (it->first == 1) {
      update(it->first, 1);
    } else {
      auto it2 = s.lower_bound(make_pair(it->first - 1, -1));
      if (it2->second > it->second) {
        update(it->first, 1);
      }
    }
  }
  cin >> q;
  for (int i = 0; i < q; ++i) {
    cin >> op >> x >> y;
    if (op == 1) {
      int ans = query(y) - query(x - 1);
      if (query(x) - query(x - 1) == 0) ans++;
      cout << ans << "\n";
    } else {
      {
        s.erase(make_pair(v[x], x));
        s.insert(make_pair(v[x], y));
        if (v[x] > 1) {
          auto it = s.lower_bound(make_pair(v[x] - 1, -1));
          if (it->second > y) {
            if (query(v[x]) - query(v[x] - 1) == 0) update(v[x], 1);
          } else {
            if (query(v[x]) - query(v[x] - 1) == 1) update(v[x], -1);
          }
        }
      }
      {
        if (v[x] < n) {
          auto it = s.lower_bound(make_pair(v[x] + 1, -1));
          if (it->second < y) {
            if (query(v[x] + 1) - query(v[x]) == 0) update(v[x] + 1, 1);
          } else {
            if (query(v[x] + 1) - query(v[x]) == 1) update(v[x] + 1, -1);
          }
        }
      }
      {
        s.erase(make_pair(v[y], y));
        s.insert(make_pair(v[y], x));
        if (v[y] > 1) {
          auto it = s.lower_bound(make_pair(v[y] - 1, -1));
          if (it->second > x) {
            if (query(v[y]) - query(v[y] - 1) == 0) update(v[y], 1);
          } else {
            if (query(v[y]) - query(v[y] - 1) == 1) update(v[y], -1);
          }
        }
      }
      {
        if (v[y] < n) {
          auto it = s.lower_bound(make_pair(v[y] + 1, -1));
          if (it->second < x) {
            if (query(v[y] + 1) - query(v[y]) == 0) update(v[y] + 1, 1);
          } else {
            if (query(v[y] + 1) - query(v[y]) == 1) update(v[y] + 1, -1);
          }
        }
      }
      swap(v[x], v[y]);
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
int n, q, op;
set<pair<int, int> > s;
int x, y;
int bit[300010];
int v[300010];
void update(int x, int v) {
  while (x < 300010) {
    bit[x] += v;
    x += (x & -x);
  }
}
int query(int x) {
  int sum = 0;
  while (x > 0) {
    sum += bit[x];
    x -= (x & -x);
  }
  return sum;
}
void printset() {
  printf("Set -------- \n");
  for (auto it = s.begin(); it != s.end(); ++it) {
    printf("%d - %d\n", it->first, it->second);
  }
}
int main(void) {
  ios ::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> v[i];
    s.insert(make_pair(v[i], i));
  }
  for (auto it = s.begin(); it != s.end(); ++it) {
    if (it->first == 1) {
      update(it->first, 1);
    } else {
      auto it2 = s.lower_bound(make_pair(it->first - 1, -1));
      if (it2->second > it->second) {
        update(it->first, 1);
      }
    }
  }
  cin >> q;
  for (int i = 0; i < q; ++i) {
    cin >> op >> x >> y;
    if (op == 1) {
      int ans = query(y) - query(x - 1);
      if (query(x) - query(x - 1) == 0) ans++;
      cout << ans << "\n";
    } else {
      {
        s.erase(make_pair(v[x], x));
        s.insert(make_pair(v[x], y));
        if (v[x] > 1) {
          auto it = s.lower_bound(make_pair(v[x] - 1, -1));
          if (it->second > y) {
            if (query(v[x]) - query(v[x] - 1) == 0) update(v[x], 1);
          } else {
            if (query(v[x]) - query(v[x] - 1) == 1) update(v[x], -1);
          }
        }
      }
      {
        if (v[x] < n) {
          auto it = s.lower_bound(make_pair(v[x] + 1, -1));
          if (it->second < y) {
            if (query(v[x] + 1) - query(v[x]) == 0) update(v[x] + 1, 1);
          } else {
            if (query(v[x] + 1) - query(v[x]) == 1) update(v[x] + 1, -1);
          }
        }
      }
      {
        s.erase(make_pair(v[y], y));
        s.insert(make_pair(v[y], x));
        if (v[y] > 1) {
          auto it = s.lower_bound(make_pair(v[y] - 1, -1));
          if (it->second > x) {
            if (query(v[y]) - query(v[y] - 1) == 0) update(v[y], 1);
          } else {
            if (query(v[y]) - query(v[y] - 1) == 1) update(v[y], -1);
          }
        }
      }
      {
        if (v[y] < n) {
          auto it = s.lower_bound(make_pair(v[y] + 1, -1));
          if (it->second < x) {
            if (query(v[y] + 1) - query(v[y]) == 0) update(v[y] + 1, 1);
          } else {
            if (query(v[y] + 1) - query(v[y]) == 1) update(v[y] + 1, -1);
          }
        }
      }
      swap(v[x], v[y]);
    }
  }
  return 0;
}

```