#include <bits/stdc++.h>
using namespace std;
int N, min_tree[500900], ar[300090], max_tree[1200900], pos[300090];
int update_max_tree(int node, int beg, int end, int L, int V) {
  if (beg > L || end < L) {
    return max_tree[node];
  }
  if (beg == end && beg == L) {
    max_tree[node] = V;
    return max_tree[node];
  }
  int P = update_max_tree(node * 2, beg, (beg + end) / 2, L, V);
  int Q = update_max_tree(node * 2 + 1, (beg + end) / 2 + 1, end, L, V);
  max_tree[node] = P + Q;
  return max_tree[node];
}
int query_tree(int node, int beg, int end, int L, int R) {
  if (beg > R || end < L) {
    return 0;
  }
  if (beg >= L && end <= R) {
    return max_tree[node];
  }
  int P = query_tree(node * 2, beg, (beg + end) / 2, L, R);
  int Q = query_tree(node * 2 + 1, (beg + end) / 2 + 1, end, L, R);
  return P + Q;
}
int main() {
  while (cin >> N) {
    memset(max_tree, 0, sizeof(max_tree));
    for (int i = 0, _n = N; i < _n; i++) {
      scanf("%d", &ar[i]);
      pos[ar[i]] = i;
    }
    for (int i = 2, _n = N + 1; i < _n; i++) {
      if (pos[i] < pos[i - 1]) {
        update_max_tree(1, 0, N, i, 1);
      }
    }
    int Q;
    scanf("%d", &Q);
    while (Q--) {
      int T, x, y;
      cin >> T >> x >> y;
      if (T == 1) {
        int maxi = query_tree(1, 0, N, x + 1, y);
        cout << maxi + 1 << endl;
      } else {
        int u = x - 1, v = y - 1;
        x = ar[x - 1];
        y = ar[y - 1];
        swap(ar[u], ar[v]);
        int newpx = pos[y];
        int newpy = pos[x];
        pos[x] = newpx;
        pos[y] = newpy;
        if (pos[x - 1] > pos[x]) {
          update_max_tree(1, 0, N, x, 1);
        } else {
          update_max_tree(1, 0, N, x, 0);
        }
        if (x + 1 <= N && pos[x + 1] < pos[x]) {
          update_max_tree(1, 0, N, x + 1, 1);
        } else {
          update_max_tree(1, 0, N, x + 1, 0);
        }
        if (pos[y - 1] > pos[y]) {
          update_max_tree(1, 0, N, y, 1);
        } else {
          update_max_tree(1, 0, N, y, 0);
        }
        if (y + 1 <= N && pos[y + 1] < pos[y]) {
          update_max_tree(1, 0, N, y + 1, 1);
        } else {
          update_max_tree(1, 0, N, y + 1, 0);
        }
      }
    }
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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, min_tree[500900], ar[300090], max_tree[1200900], pos[300090];
int update_max_tree(int node, int beg, int end, int L, int V) {
  if (beg > L || end < L) {
    return max_tree[node];
  }
  if (beg == end && beg == L) {
    max_tree[node] = V;
    return max_tree[node];
  }
  int P = update_max_tree(node * 2, beg, (beg + end) / 2, L, V);
  int Q = update_max_tree(node * 2 + 1, (beg + end) / 2 + 1, end, L, V);
  max_tree[node] = P + Q;
  return max_tree[node];
}
int query_tree(int node, int beg, int end, int L, int R) {
  if (beg > R || end < L) {
    return 0;
  }
  if (beg >= L && end <= R) {
    return max_tree[node];
  }
  int P = query_tree(node * 2, beg, (beg + end) / 2, L, R);
  int Q = query_tree(node * 2 + 1, (beg + end) / 2 + 1, end, L, R);
  return P + Q;
}
int main() {
  while (cin >> N) {
    memset(max_tree, 0, sizeof(max_tree));
    for (int i = 0, _n = N; i < _n; i++) {
      scanf("%d", &ar[i]);
      pos[ar[i]] = i;
    }
    for (int i = 2, _n = N + 1; i < _n; i++) {
      if (pos[i] < pos[i - 1]) {
        update_max_tree(1, 0, N, i, 1);
      }
    }
    int Q;
    scanf("%d", &Q);
    while (Q--) {
      int T, x, y;
      cin >> T >> x >> y;
      if (T == 1) {
        int maxi = query_tree(1, 0, N, x + 1, y);
        cout << maxi + 1 << endl;
      } else {
        int u = x - 1, v = y - 1;
        x = ar[x - 1];
        y = ar[y - 1];
        swap(ar[u], ar[v]);
        int newpx = pos[y];
        int newpy = pos[x];
        pos[x] = newpx;
        pos[y] = newpy;
        if (pos[x - 1] > pos[x]) {
          update_max_tree(1, 0, N, x, 1);
        } else {
          update_max_tree(1, 0, N, x, 0);
        }
        if (x + 1 <= N && pos[x + 1] < pos[x]) {
          update_max_tree(1, 0, N, x + 1, 1);
        } else {
          update_max_tree(1, 0, N, x + 1, 0);
        }
        if (pos[y - 1] > pos[y]) {
          update_max_tree(1, 0, N, y, 1);
        } else {
          update_max_tree(1, 0, N, y, 0);
        }
        if (y + 1 <= N && pos[y + 1] < pos[y]) {
          update_max_tree(1, 0, N, y + 1, 1);
        } else {
          update_max_tree(1, 0, N, y + 1, 0);
        }
      }
    }
  }
  return 0;
}

```