#include <bits/stdc++.h>
using namespace std;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
int s[300009 << 2], pos[300009], arr[300009];
pair<int, int> range[300009 << 2];
vector<int> adj[300009];
void build(int nd, int x, int y) {
  if (x == y) {
    s[nd] = 1;
    return;
  }
  int mid = (x + y) >> 1;
  build(nd << 1, x, mid);
  build(nd << 1 | 1, mid + 1, y);
  s[nd] = s[nd << 1] + s[nd << 1 | 1];
  if (pos[mid] < pos[mid + 1]) s[nd]--;
  adj[mid].push_back(nd);
  adj[mid + 1].push_back(nd);
  range[nd] = make_pair(x, y);
}
void upd(int l, int r, int id, int nd, int x, int y) {
  if (l > y or x > r or nd > id) return;
  int mid = (x + y) >> 1;
  upd(l, r, id, nd << 1, x, mid);
  upd(l, r, id, nd << 1 | 1, mid + 1, y);
  s[nd] = s[nd << 1] + s[nd << 1 | 1];
  if (pos[mid] < pos[mid + 1]) s[nd]--;
}
int tap(int l, int r, int nd, int x, int y) {
  if (l > y or x > r) return -1;
  if (l <= x and y <= r) return s[nd];
  int mid = (x + y) >> 1;
  int i1 = tap(l, r, nd << 1, x, mid);
  int i2 = tap(l, r, nd << 1 | 1, mid + 1, y);
  if (i1 == -1) return i2;
  if (i2 == -1) return i1;
  if (pos[mid] < pos[mid + 1]) return i1 + i2 - 1;
  return i1 + i2;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", arr + i), pos[arr[i]] = i;
  build(1, 1, n);
  int q;
  scanf("%d", &q);
  while (q--) {
    int type, l, r;
    scanf("%d%d%d", &type, &l, &r);
    if (type == 2) {
      pos[arr[l]] = r;
      pos[arr[r]] = l;
      swap(arr[l], arr[r]);
      for (__typeof((adj[arr[l]]).begin()) it = (adj[arr[l]]).begin();
           it != (adj[arr[l]]).end(); it++)
        upd(range[*it].first, range[*it].second, *it, 1, 1, n);
      for (__typeof((adj[arr[r]]).begin()) it = (adj[arr[r]]).begin();
           it != (adj[arr[r]]).end(); it++)
        upd(range[*it].first, range[*it].second, *it, 1, 1, n);
    } else
      printf("%d\n", tap(l, r, 1, 1, n));
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
int s[300009 << 2], pos[300009], arr[300009];
pair<int, int> range[300009 << 2];
vector<int> adj[300009];
void build(int nd, int x, int y) {
  if (x == y) {
    s[nd] = 1;
    return;
  }
  int mid = (x + y) >> 1;
  build(nd << 1, x, mid);
  build(nd << 1 | 1, mid + 1, y);
  s[nd] = s[nd << 1] + s[nd << 1 | 1];
  if (pos[mid] < pos[mid + 1]) s[nd]--;
  adj[mid].push_back(nd);
  adj[mid + 1].push_back(nd);
  range[nd] = make_pair(x, y);
}
void upd(int l, int r, int id, int nd, int x, int y) {
  if (l > y or x > r or nd > id) return;
  int mid = (x + y) >> 1;
  upd(l, r, id, nd << 1, x, mid);
  upd(l, r, id, nd << 1 | 1, mid + 1, y);
  s[nd] = s[nd << 1] + s[nd << 1 | 1];
  if (pos[mid] < pos[mid + 1]) s[nd]--;
}
int tap(int l, int r, int nd, int x, int y) {
  if (l > y or x > r) return -1;
  if (l <= x and y <= r) return s[nd];
  int mid = (x + y) >> 1;
  int i1 = tap(l, r, nd << 1, x, mid);
  int i2 = tap(l, r, nd << 1 | 1, mid + 1, y);
  if (i1 == -1) return i2;
  if (i2 == -1) return i1;
  if (pos[mid] < pos[mid + 1]) return i1 + i2 - 1;
  return i1 + i2;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", arr + i), pos[arr[i]] = i;
  build(1, 1, n);
  int q;
  scanf("%d", &q);
  while (q--) {
    int type, l, r;
    scanf("%d%d%d", &type, &l, &r);
    if (type == 2) {
      pos[arr[l]] = r;
      pos[arr[r]] = l;
      swap(arr[l], arr[r]);
      for (__typeof((adj[arr[l]]).begin()) it = (adj[arr[l]]).begin();
           it != (adj[arr[l]]).end(); it++)
        upd(range[*it].first, range[*it].second, *it, 1, 1, n);
      for (__typeof((adj[arr[r]]).begin()) it = (adj[arr[r]]).begin();
           it != (adj[arr[r]]).end(); it++)
        upd(range[*it].first, range[*it].second, *it, 1, 1, n);
    } else
      printf("%d\n", tap(l, r, 1, 1, n));
  }
  return 0;
}

```