#include <bits/stdc++.h>
using namespace std;
struct debugger {
  static void call(string::iterator it, string::iterator ed) {}
  template <typename T, typename... aT>
  static void call(string::iterator it, string::iterator ed, T a, aT... rest) {
    string b;
    for (; *it != ','; ++it)
      if (*it != ' ') b += *it;
    cerr << b << " = " << a << " ";
    call(++it, ed, rest...);
  }
};
const int MAX = 300001;
struct node {
  int mn, mx, r;
  node() {
    mn = mx = -1;
    r = 1;
  }
};
node tr[MAX * 4 + 7];
int ar[MAX + 7];
void update(int p, int l, int r, int pos, int val) {
  if (l == r) {
    tr[p].mn = val;
    tr[p].mx = val;
    tr[p].r = 1;
    return;
  }
  int m = (l + r) / 2;
  if (pos <= m)
    update(p * 2 + 1, l, m, pos, val);
  else
    update(p * 2 + 2, m + 1, r, pos, val);
  tr[p].r = tr[p * 2 + 1].r + tr[p * 2 + 2].r;
  if (tr[p * 2 + 1].mx <= tr[p * 2 + 2].mn) {
    tr[p].r--;
  }
  tr[p].mn = tr[p * 2 + 1].mn;
  tr[p].mx = tr[p * 2 + 2].mx;
}
bool qu(int p, int L, int R, int l, int r, node &s) {
  if (L >= l && R <= r) {
    s = tr[p];
    return true;
  }
  if (L > r || R < l) {
    return false;
  }
  int m = (L + R) / 2;
  node e, f, g;
  bool a = qu(p * 2 + 1, L, m, l, r, e);
  bool b = qu(p * 2 + 2, m + 1, R, l, r, f);
  if (a && b) {
    g.r = e.r + f.r;
    if (e.mx <= f.mn) {
      g.r--;
    }
    g.mn = e.mn;
    g.mx = f.mx;
    s = g;
  } else if (a) {
    s = e;
  } else if (b) {
    s = f;
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ar[i];
    update(0, 1, n, ar[i], i);
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int p, x, y;
    cin >> p >> x >> y;
    if (p == 1) {
      node u;
      qu(0, 1, n, x, y, u);
      cout << u.r << endl;
    } else {
      update(0, 1, n, ar[x], y);
      update(0, 1, n, ar[y], x);
      swap(ar[x], ar[y]);
    }
  }
}

### Prompt

Generate 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;
struct debugger {
  static void call(string::iterator it, string::iterator ed) {}
  template <typename T, typename... aT>
  static void call(string::iterator it, string::iterator ed, T a, aT... rest) {
    string b;
    for (; *it != ','; ++it)
      if (*it != ' ') b += *it;
    cerr << b << " = " << a << " ";
    call(++it, ed, rest...);
  }
};
const int MAX = 300001;
struct node {
  int mn, mx, r;
  node() {
    mn = mx = -1;
    r = 1;
  }
};
node tr[MAX * 4 + 7];
int ar[MAX + 7];
void update(int p, int l, int r, int pos, int val) {
  if (l == r) {
    tr[p].mn = val;
    tr[p].mx = val;
    tr[p].r = 1;
    return;
  }
  int m = (l + r) / 2;
  if (pos <= m)
    update(p * 2 + 1, l, m, pos, val);
  else
    update(p * 2 + 2, m + 1, r, pos, val);
  tr[p].r = tr[p * 2 + 1].r + tr[p * 2 + 2].r;
  if (tr[p * 2 + 1].mx <= tr[p * 2 + 2].mn) {
    tr[p].r--;
  }
  tr[p].mn = tr[p * 2 + 1].mn;
  tr[p].mx = tr[p * 2 + 2].mx;
}
bool qu(int p, int L, int R, int l, int r, node &s) {
  if (L >= l && R <= r) {
    s = tr[p];
    return true;
  }
  if (L > r || R < l) {
    return false;
  }
  int m = (L + R) / 2;
  node e, f, g;
  bool a = qu(p * 2 + 1, L, m, l, r, e);
  bool b = qu(p * 2 + 2, m + 1, R, l, r, f);
  if (a && b) {
    g.r = e.r + f.r;
    if (e.mx <= f.mn) {
      g.r--;
    }
    g.mn = e.mn;
    g.mx = f.mx;
    s = g;
  } else if (a) {
    s = e;
  } else if (b) {
    s = f;
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ar[i];
    update(0, 1, n, ar[i], i);
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int p, x, y;
    cin >> p >> x >> y;
    if (p == 1) {
      node u;
      qu(0, 1, n, x, y, u);
      cout << u.r << endl;
    } else {
      update(0, 1, n, ar[x], y);
      update(0, 1, n, ar[y], x);
      swap(ar[x], ar[y]);
    }
  }
}

```