#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 300000;
const int MAX_E2 = 1 << 20;
struct Elm {
  int l, r, v;
  Elm() {}
  Elm(int _l, int _r, int _v) : l(_l), r(_r), v(_v) {}
  Elm(int _l) : l(_l), r(_l), v(1) {}
  Elm operator+(const Elm &e) const {
    if (v == 0) return e;
    if (e.v == 0) return *this;
    Elm a;
    a.l = l, a.r = e.r;
    a.v = v + e.v - ((r < e.l) ? 1 : 0);
    return a;
  }
};
template <typename T, const int MAX_E2>
struct SegTreeSum {
  int n, e2;
  T nodes[MAX_E2], z;
  SegTreeSum() {}
  void init(int n, T _z) {
    z = _z;
    for (e2 = 1; e2 < n; e2 <<= 1)
      ;
    fill(nodes, nodes + MAX_E2, z);
  }
  T &get(int i) { return nodes[e2 - 1 + i]; }
  void setall() {
    for (int j = e2 - 2; j >= 0; j--)
      nodes[j] = nodes[j * 2 + 1] + nodes[j * 2 + 2];
  }
  void set(int i, T v) {
    int j = e2 - 1 + i;
    nodes[j] = v;
    while (j > 0) {
      j = (j - 1) / 2;
      nodes[j] = nodes[j * 2 + 1] + nodes[j * 2 + 2];
    }
  }
  T sum_range(int r0, int r1, int k, int i0, int i1) {
    if (r1 <= i0 || i1 <= r0) return z;
    if (r0 <= i0 && i1 <= r1) return nodes[k];
    int im = (i0 + i1) / 2;
    T v0 = sum_range(r0, r1, k * 2 + 1, i0, im);
    T v1 = sum_range(r0, r1, k * 2 + 2, im, i1);
    return v0 + v1;
  }
  T sum_range(int r0, int r1) { return sum_range(r0, r1, 0, 0, e2); }
};
int as[MAX_N];
SegTreeSum<Elm, MAX_E2> st;
int main() {
  int n;
  scanf("%d", &n);
  st.init(n, Elm(0, 0, 0));
  for (int i = 0; i < n; i++) {
    scanf("%d", as + i);
    as[i]--;
    st.get(as[i]) = Elm(i, i, 1);
  }
  st.setall();
  int q;
  scanf("%d", &q);
  while (q--) {
    int p, x, y;
    scanf("%d%d%d", &p, &x, &y);
    if (p == 1) {
      x--;
      Elm e = st.sum_range(x, y);
      printf("%d\n", e.v);
    } else {
      x--, y--;
      int ax = as[x], ay = as[y];
      swap(as[x], as[y]);
      st.set(ax, Elm(y));
      st.set(ay, Elm(x));
    }
  }
  return 0;
}

### 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;
const int MAX_N = 300000;
const int MAX_E2 = 1 << 20;
struct Elm {
  int l, r, v;
  Elm() {}
  Elm(int _l, int _r, int _v) : l(_l), r(_r), v(_v) {}
  Elm(int _l) : l(_l), r(_l), v(1) {}
  Elm operator+(const Elm &e) const {
    if (v == 0) return e;
    if (e.v == 0) return *this;
    Elm a;
    a.l = l, a.r = e.r;
    a.v = v + e.v - ((r < e.l) ? 1 : 0);
    return a;
  }
};
template <typename T, const int MAX_E2>
struct SegTreeSum {
  int n, e2;
  T nodes[MAX_E2], z;
  SegTreeSum() {}
  void init(int n, T _z) {
    z = _z;
    for (e2 = 1; e2 < n; e2 <<= 1)
      ;
    fill(nodes, nodes + MAX_E2, z);
  }
  T &get(int i) { return nodes[e2 - 1 + i]; }
  void setall() {
    for (int j = e2 - 2; j >= 0; j--)
      nodes[j] = nodes[j * 2 + 1] + nodes[j * 2 + 2];
  }
  void set(int i, T v) {
    int j = e2 - 1 + i;
    nodes[j] = v;
    while (j > 0) {
      j = (j - 1) / 2;
      nodes[j] = nodes[j * 2 + 1] + nodes[j * 2 + 2];
    }
  }
  T sum_range(int r0, int r1, int k, int i0, int i1) {
    if (r1 <= i0 || i1 <= r0) return z;
    if (r0 <= i0 && i1 <= r1) return nodes[k];
    int im = (i0 + i1) / 2;
    T v0 = sum_range(r0, r1, k * 2 + 1, i0, im);
    T v1 = sum_range(r0, r1, k * 2 + 2, im, i1);
    return v0 + v1;
  }
  T sum_range(int r0, int r1) { return sum_range(r0, r1, 0, 0, e2); }
};
int as[MAX_N];
SegTreeSum<Elm, MAX_E2> st;
int main() {
  int n;
  scanf("%d", &n);
  st.init(n, Elm(0, 0, 0));
  for (int i = 0; i < n; i++) {
    scanf("%d", as + i);
    as[i]--;
    st.get(as[i]) = Elm(i, i, 1);
  }
  st.setall();
  int q;
  scanf("%d", &q);
  while (q--) {
    int p, x, y;
    scanf("%d%d%d", &p, &x, &y);
    if (p == 1) {
      x--;
      Elm e = st.sum_range(x, y);
      printf("%d\n", e.v);
    } else {
      x--, y--;
      int ax = as[x], ay = as[y];
      swap(as[x], as[y]);
      st.set(ax, Elm(y));
      st.set(ay, Elm(x));
    }
  }
  return 0;
}

```