#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-9;
const int MAXN = 3 * int(1e5) + 10;
int pos[MAXN];
int a[MAXN];
int n, q;
int deleted = -1;
int fenw[MAXN];
void upd(int x, int d) {
  for (int i = x; i < n; i = (i | (i + 1))) fenw[i] += d;
}
int sum(int r) {
  int res = 0;
  for (int i = r; i >= 0; i = (i & (i + 1)) - 1) res += fenw[i];
  return res;
}
int sum(int l, int r) { return sum(r) - sum(l - 1); }
bool isBeg(int i) { return sum(i, i) == 1; }
void build() {
  memset(fenw, 0, sizeof fenw);
  upd(0, 1);
  for (int i = 1; i < n; i++)
    if (pos[i - 1] > pos[i]) upd(i, 1);
}
void del(int val) {
  if (val < n - 1 && pos[val] < pos[val + 1] && deleted != val + 1)
    upd(val + 1, 1);
  if (isBeg(val)) upd(val, -1);
}
void add(int val) {
  if (val < n - 1 && pos[val] < pos[val + 1] && deleted != val + 1)
    upd(val + 1, -1);
  if (0 == val || pos[val - 1] > pos[val] || deleted == val - 1) upd(val, 1);
}
void goSwap(int x, int y) {
  deleted = -1;
  del(a[x]);
  deleted = a[x];
  del(a[y]);
  swap(a[x], a[y]);
  pos[a[x]] = x;
  pos[a[y]] = y;
  add(a[x]);
  add(a[y]);
}
int calc(int x, int y) {
  if (isBeg(x))
    return sum(x, y);
  else
    return sum(x, y) + 1;
}
void output() {
  for (int i = 0; i < int(n); i++) printf("%d ", sum(i, i));
  puts("");
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < int(n); i++) {
    scanf("%d", a + i);
    a[i]--;
    pos[a[i]] = i;
  }
  build();
  scanf("%d", &q);
  for (int i = 0; i < int(q); i++) {
    int p, x, y;
    scanf("%d %d %d", &p, &x, &y);
    x--;
    y--;
    if (p == 1)
      printf("%d\n", calc(x, y));
    else
      goSwap(x, y);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
const double EPS = 1e-9;
const int MAXN = 3 * int(1e5) + 10;
int pos[MAXN];
int a[MAXN];
int n, q;
int deleted = -1;
int fenw[MAXN];
void upd(int x, int d) {
  for (int i = x; i < n; i = (i | (i + 1))) fenw[i] += d;
}
int sum(int r) {
  int res = 0;
  for (int i = r; i >= 0; i = (i & (i + 1)) - 1) res += fenw[i];
  return res;
}
int sum(int l, int r) { return sum(r) - sum(l - 1); }
bool isBeg(int i) { return sum(i, i) == 1; }
void build() {
  memset(fenw, 0, sizeof fenw);
  upd(0, 1);
  for (int i = 1; i < n; i++)
    if (pos[i - 1] > pos[i]) upd(i, 1);
}
void del(int val) {
  if (val < n - 1 && pos[val] < pos[val + 1] && deleted != val + 1)
    upd(val + 1, 1);
  if (isBeg(val)) upd(val, -1);
}
void add(int val) {
  if (val < n - 1 && pos[val] < pos[val + 1] && deleted != val + 1)
    upd(val + 1, -1);
  if (0 == val || pos[val - 1] > pos[val] || deleted == val - 1) upd(val, 1);
}
void goSwap(int x, int y) {
  deleted = -1;
  del(a[x]);
  deleted = a[x];
  del(a[y]);
  swap(a[x], a[y]);
  pos[a[x]] = x;
  pos[a[y]] = y;
  add(a[x]);
  add(a[y]);
}
int calc(int x, int y) {
  if (isBeg(x))
    return sum(x, y);
  else
    return sum(x, y) + 1;
}
void output() {
  for (int i = 0; i < int(n); i++) printf("%d ", sum(i, i));
  puts("");
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < int(n); i++) {
    scanf("%d", a + i);
    a[i]--;
    pos[a[i]] = i;
  }
  build();
  scanf("%d", &q);
  for (int i = 0; i < int(q); i++) {
    int p, x, y;
    scanf("%d %d %d", &p, &x, &y);
    x--;
    y--;
    if (p == 1)
      printf("%d\n", calc(x, y));
    else
      goSwap(x, y);
  }
  return 0;
}

```