#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:64000000")
const int maxn = 1 << 19;
const int inf = 1000000007;
const int mod = 1000000007;
int n, m;
vector<int> a, b;
int tp[maxn], x[maxn], y[maxn];
void reading() {
  scanf("%d", &n);
  a.resize(n + 1);
  b.resize(n + 1);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[a[i]] = i;
  }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &tp[i], &x[i], &y[i]);
  }
}
void print(vector<int> v) {
  for (int i = 0; i < v.size(); i++) printf("%d\n", v[i]);
}
int calcdp(int l, int r) {
  int res = 1;
  for (int i = l; i < r; i++) {
    if (b[i + 1] < b[i]) res++;
  }
  return res;
}
int t[2 * maxn], tl[2 * maxn], tm[2 * maxn], tr[2 * maxn];
int combine(int v, int x, int u, int y) { return v + u - (x < y); }
void build(int v, int TL, int TR) {
  tl[v] = TL;
  tr[v] = TR;
  if (tl[v] == tr[v]) {
    t[v] = 1;
    return;
  }
  tm[v] = (tl[v] + tr[v]) >> 1;
  build(v * 2, tl[v], tm[v]);
  build(v * 2 + 1, tm[v] + 1, tr[v]);
  t[v] = combine(t[v * 2], b[tr[v * 2]], t[v * 2 + 1], b[tl[v * 2 + 1]]);
}
int get(int v, int l, int r) {
  if (l > r) return 0;
  if (tl[v] == l && tr[v] == r) return t[v];
  int o1 = get(v * 2, l, min(r, tm[v]));
  int o2 = get(v * 2 + 1, max(l, tm[v] + 1), r);
  return combine(o1, (o1 > 0 ? b[tr[v * 2]] : inf), o2,
                 (o2 > 0 ? b[tl[v * 2 + 1]] : -inf));
}
void upd(int v, int x, int y) {
  if (tl[v] == tr[v]) {
    b[tl[v]] = y;
    return;
  }
  if (x <= tm[v])
    upd(v * 2, x, y);
  else
    upd(v * 2 + 1, x, y);
  t[v] = combine(t[v * 2], b[tr[v * 2]], t[v * 2 + 1], b[tl[v * 2 + 1]]);
}
vector<int> slowsolve() {
  vector<int> res;
  build(1, 0, n);
  for (int i = 0; i < m; i++) {
    if (tp[i] == 1) {
      res.push_back(get(1, x[i], y[i]));
    }
    if (tp[i] == 2) {
      upd(1, a[x[i]], y[i]);
      upd(1, a[y[i]], x[i]);
      swap(a[x[i]], a[y[i]]);
    }
  }
  for (int i = m - 1; i >= 0; i--)
    if (tp[i] == 2) {
      swap(a[x[i]], a[y[i]]);
      b[a[x[i]]] = x[i];
      b[a[y[i]]] = y[i];
    }
  return res;
}
int main() {
  reading();
  print(slowsolve());
  return 0;
}

### 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;
#pragma comment(linker, "/STACK:64000000")
const int maxn = 1 << 19;
const int inf = 1000000007;
const int mod = 1000000007;
int n, m;
vector<int> a, b;
int tp[maxn], x[maxn], y[maxn];
void reading() {
  scanf("%d", &n);
  a.resize(n + 1);
  b.resize(n + 1);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[a[i]] = i;
  }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &tp[i], &x[i], &y[i]);
  }
}
void print(vector<int> v) {
  for (int i = 0; i < v.size(); i++) printf("%d\n", v[i]);
}
int calcdp(int l, int r) {
  int res = 1;
  for (int i = l; i < r; i++) {
    if (b[i + 1] < b[i]) res++;
  }
  return res;
}
int t[2 * maxn], tl[2 * maxn], tm[2 * maxn], tr[2 * maxn];
int combine(int v, int x, int u, int y) { return v + u - (x < y); }
void build(int v, int TL, int TR) {
  tl[v] = TL;
  tr[v] = TR;
  if (tl[v] == tr[v]) {
    t[v] = 1;
    return;
  }
  tm[v] = (tl[v] + tr[v]) >> 1;
  build(v * 2, tl[v], tm[v]);
  build(v * 2 + 1, tm[v] + 1, tr[v]);
  t[v] = combine(t[v * 2], b[tr[v * 2]], t[v * 2 + 1], b[tl[v * 2 + 1]]);
}
int get(int v, int l, int r) {
  if (l > r) return 0;
  if (tl[v] == l && tr[v] == r) return t[v];
  int o1 = get(v * 2, l, min(r, tm[v]));
  int o2 = get(v * 2 + 1, max(l, tm[v] + 1), r);
  return combine(o1, (o1 > 0 ? b[tr[v * 2]] : inf), o2,
                 (o2 > 0 ? b[tl[v * 2 + 1]] : -inf));
}
void upd(int v, int x, int y) {
  if (tl[v] == tr[v]) {
    b[tl[v]] = y;
    return;
  }
  if (x <= tm[v])
    upd(v * 2, x, y);
  else
    upd(v * 2 + 1, x, y);
  t[v] = combine(t[v * 2], b[tr[v * 2]], t[v * 2 + 1], b[tl[v * 2 + 1]]);
}
vector<int> slowsolve() {
  vector<int> res;
  build(1, 0, n);
  for (int i = 0; i < m; i++) {
    if (tp[i] == 1) {
      res.push_back(get(1, x[i], y[i]));
    }
    if (tp[i] == 2) {
      upd(1, a[x[i]], y[i]);
      upd(1, a[y[i]], x[i]);
      swap(a[x[i]], a[y[i]]);
    }
  }
  for (int i = m - 1; i >= 0; i--)
    if (tp[i] == 2) {
      swap(a[x[i]], a[y[i]]);
      b[a[x[i]]] = x[i];
      b[a[y[i]]] = y[i];
    }
  return res;
}
int main() {
  reading();
  print(slowsolve());
  return 0;
}

```