#include <bits/stdc++.h>
std::ofstream fout("out.txt");
int sum[4 * 300005];
void update(int node, int left, int right, int pos, int val) {
  if (left > pos or right < pos) return;
  if (left == right) {
    sum[node] = val;
    return;
  }
  int mid = (left + right) / 2;
  update(2 * node + 1, left, mid, pos, val);
  update(2 * node + 2, mid + 1, right, pos, val);
  sum[node] = sum[2 * node + 1] + sum[2 * node + 2];
}
int query(int node, int left, int right, int l, int r) {
  if (left > r or right < l) return 0;
  if (l <= left and right <= r) return sum[node];
  int mid = (left + right) / 2;
  return query(2 * node + 1, left, mid, l, r) +
         query(2 * node + 2, mid + 1, right, l, r);
}
int pos[300005];
int x[300005];
void print(int node, int left, int right) {
  if (left == right) {
    std::cout << sum[node] << ' ';
    return;
  }
  int mid = (left + right) / 2;
  print(2 * node + 1, left, mid);
  print(2 * node + 2, mid + 1, right);
}
int bit[300005];
void add(int n, int pos, int val) {
  for (; pos <= n; pos = (pos | (pos + 1))) bit[pos] += val;
}
int q(int n, int pos) {
  int ans = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) ans += bit[pos];
  return ans;
}
bool ok[300005];
int main() {
  int n;
  std::cin >> n;
  for (int i = 1; i <= n; i++) {
    std::cin >> x[i];
    pos[x[i]] = i;
  }
  for (int i = 1; i < n; i++) {
    if (pos[i] > pos[i + 1]) {
      add(n, i + 1, 1);
      ok[i + 1] = 1;
    }
  }
  int Q;
  std::cin >> Q;
  while (Q--) {
    int type, a, b;
    std::cin >> type >> a >> b;
    if (type == 1) {
      std::cout << q(n, b) - q(n, a) + 1 << '\n';
    } else {
      std::swap(x[a], x[b]);
      pos[x[a]] = a;
      pos[x[b]] = b;
      a = x[a];
      b = x[b];
      if (a > 1 and ok[a] != (pos[a - 1] > pos[a])) {
        if (ok[a] == 0)
          add(n, a, 1);
        else
          add(n, a, -1);
        ok[a] = 1 ^ ok[a];
      }
      if (a < n and ok[a + 1] != (pos[a] > pos[a + 1])) {
        if (ok[a + 1] == 0)
          add(n, a + 1, 1);
        else
          add(n, a + 1, -1);
        ok[a + 1] = 1 ^ ok[a + 1];
      }
      if (b > 1 and ok[b] != (pos[b - 1] > pos[b])) {
        if (ok[b] == 0)
          add(n, b, 1);
        else
          add(n, b, -1);
        ok[b] = 1 ^ ok[b];
      }
      if (b < n and ok[b + 1] != (pos[b] > pos[b + 1])) {
        if (ok[b + 1] == 0)
          add(n, b + 1, 1);
        else
          add(n, b + 1, -1);
        ok[b + 1] = 1 ^ ok[b + 1];
      }
    }
  }
  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>
std::ofstream fout("out.txt");
int sum[4 * 300005];
void update(int node, int left, int right, int pos, int val) {
  if (left > pos or right < pos) return;
  if (left == right) {
    sum[node] = val;
    return;
  }
  int mid = (left + right) / 2;
  update(2 * node + 1, left, mid, pos, val);
  update(2 * node + 2, mid + 1, right, pos, val);
  sum[node] = sum[2 * node + 1] + sum[2 * node + 2];
}
int query(int node, int left, int right, int l, int r) {
  if (left > r or right < l) return 0;
  if (l <= left and right <= r) return sum[node];
  int mid = (left + right) / 2;
  return query(2 * node + 1, left, mid, l, r) +
         query(2 * node + 2, mid + 1, right, l, r);
}
int pos[300005];
int x[300005];
void print(int node, int left, int right) {
  if (left == right) {
    std::cout << sum[node] << ' ';
    return;
  }
  int mid = (left + right) / 2;
  print(2 * node + 1, left, mid);
  print(2 * node + 2, mid + 1, right);
}
int bit[300005];
void add(int n, int pos, int val) {
  for (; pos <= n; pos = (pos | (pos + 1))) bit[pos] += val;
}
int q(int n, int pos) {
  int ans = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) ans += bit[pos];
  return ans;
}
bool ok[300005];
int main() {
  int n;
  std::cin >> n;
  for (int i = 1; i <= n; i++) {
    std::cin >> x[i];
    pos[x[i]] = i;
  }
  for (int i = 1; i < n; i++) {
    if (pos[i] > pos[i + 1]) {
      add(n, i + 1, 1);
      ok[i + 1] = 1;
    }
  }
  int Q;
  std::cin >> Q;
  while (Q--) {
    int type, a, b;
    std::cin >> type >> a >> b;
    if (type == 1) {
      std::cout << q(n, b) - q(n, a) + 1 << '\n';
    } else {
      std::swap(x[a], x[b]);
      pos[x[a]] = a;
      pos[x[b]] = b;
      a = x[a];
      b = x[b];
      if (a > 1 and ok[a] != (pos[a - 1] > pos[a])) {
        if (ok[a] == 0)
          add(n, a, 1);
        else
          add(n, a, -1);
        ok[a] = 1 ^ ok[a];
      }
      if (a < n and ok[a + 1] != (pos[a] > pos[a + 1])) {
        if (ok[a + 1] == 0)
          add(n, a + 1, 1);
        else
          add(n, a + 1, -1);
        ok[a + 1] = 1 ^ ok[a + 1];
      }
      if (b > 1 and ok[b] != (pos[b - 1] > pos[b])) {
        if (ok[b] == 0)
          add(n, b, 1);
        else
          add(n, b, -1);
        ok[b] = 1 ^ ok[b];
      }
      if (b < n and ok[b + 1] != (pos[b] > pos[b + 1])) {
        if (ok[b + 1] == 0)
          add(n, b + 1, 1);
        else
          add(n, b + 1, -1);
        ok[b + 1] = 1 ^ ok[b + 1];
      }
    }
  }
  return 0;
}

```