#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 15;
int sum[N];
int inverse[N];
void update(int *pen, int ind, int add) {
  while (ind < N) pen[ind] += add, ind += ind & (-ind);
}
int query(int *pen, int ind) {
  int ret = 0;
  while (ind) ret += pen[ind], ind = ind & (ind - 1);
  return ret;
}
int cur[N];
int a[N];
int get(int x, int n) {
  int l = x, r = n, mid;
  int u = query(inverse, x - 1), v;
  while (l < r) {
    mid = (l + r) / 2;
    v = query(inverse, mid);
    if (v - u > 0) {
      r = mid;
    } else
      l = mid + 1;
  }
  return l;
}
void INupdate(int u, int v, int n, int add) {
  set<int> s;
  set<int>::iterator it;
  if (u - 1 > 0 && cur[u - 1] > cur[u]) s.insert(u - 1);
  if (u + 1 <= n && cur[u] > cur[u + 1]) s.insert(u);
  if (v - 1 > 0 && cur[v - 1] > cur[v]) s.insert(v - 1);
  if (v + 1 <= n && cur[v] > cur[v + 1]) s.insert(v);
  for (it = s.begin(); it != s.end(); ++it) update(inverse, *it, add);
}
int main() {
  int n, i;
  cin >> n;
  for (i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    cur[a[i]] = i;
  }
  for (i = 1; i < n; ++i) {
    if (cur[i] > cur[i + 1]) update(inverse, i, 1);
  }
  int pos = 1, npos;
  while (pos <= n) {
    npos = get(pos, n);
    update(sum, npos, 1);
    pos = npos + 1;
  }
  int q, type, x, y, t, nt, ind;
  cin >> q;
  set<int> s;
  set<int>::iterator it;
  while (q--) {
    scanf("%d%d%d", &type, &x, &y);
    if (type == 1) {
      pos = get(y, n);
      printf("%d\n", query(inverse, y - 1) - query(inverse, x - 1) + 1);
    } else {
      swap(a[x], a[y]);
      int u = a[x], v = a[y];
      INupdate(u, v, n, -1);
      cur[a[x]] = x, cur[a[y]] = y;
      INupdate(u, v, n, 1);
    }
  }
}

### Prompt

Please create a solution in Cpp 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;
const int N = 3e5 + 15;
int sum[N];
int inverse[N];
void update(int *pen, int ind, int add) {
  while (ind < N) pen[ind] += add, ind += ind & (-ind);
}
int query(int *pen, int ind) {
  int ret = 0;
  while (ind) ret += pen[ind], ind = ind & (ind - 1);
  return ret;
}
int cur[N];
int a[N];
int get(int x, int n) {
  int l = x, r = n, mid;
  int u = query(inverse, x - 1), v;
  while (l < r) {
    mid = (l + r) / 2;
    v = query(inverse, mid);
    if (v - u > 0) {
      r = mid;
    } else
      l = mid + 1;
  }
  return l;
}
void INupdate(int u, int v, int n, int add) {
  set<int> s;
  set<int>::iterator it;
  if (u - 1 > 0 && cur[u - 1] > cur[u]) s.insert(u - 1);
  if (u + 1 <= n && cur[u] > cur[u + 1]) s.insert(u);
  if (v - 1 > 0 && cur[v - 1] > cur[v]) s.insert(v - 1);
  if (v + 1 <= n && cur[v] > cur[v + 1]) s.insert(v);
  for (it = s.begin(); it != s.end(); ++it) update(inverse, *it, add);
}
int main() {
  int n, i;
  cin >> n;
  for (i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    cur[a[i]] = i;
  }
  for (i = 1; i < n; ++i) {
    if (cur[i] > cur[i + 1]) update(inverse, i, 1);
  }
  int pos = 1, npos;
  while (pos <= n) {
    npos = get(pos, n);
    update(sum, npos, 1);
    pos = npos + 1;
  }
  int q, type, x, y, t, nt, ind;
  cin >> q;
  set<int> s;
  set<int>::iterator it;
  while (q--) {
    scanf("%d%d%d", &type, &x, &y);
    if (type == 1) {
      pos = get(y, n);
      printf("%d\n", query(inverse, y - 1) - query(inverse, x - 1) + 1);
    } else {
      swap(a[x], a[y]);
      int u = a[x], v = a[y];
      INupdate(u, v, n, -1);
      cur[a[x]] = x, cur[a[y]] = y;
      INupdate(u, v, n, 1);
    }
  }
}

```