#include <bits/stdc++.h>
using namespace std;
const int MAXN = 300005;
int n, m;
int a[MAXN], t[MAXN], pos[MAXN];
inline int lowbit(int i) { return i & (-i); }
void add(int x, int k) {
  for (int i = x; i <= n; i += lowbit(i)) t[i] += k;
}
int get(int x) {
  int res = 0;
  for (int i = x; i; i -= lowbit(i)) res += t[i];
  return res;
}
void query(int x, int y) {
  int ans = get(y) - get(x) + 1;
  printf("%d\n", ans);
}
void change(int x, int y) {
  if (x > pos[a[x] - 1] && y <= pos[a[x] - 1]) add(a[x], 1);
  if (x < pos[a[x] - 1] && y >= pos[a[x] - 1]) add(a[x], -1);
  if (pos[a[x] + 1] > x && pos[a[x] + 1] <= y) add(a[x] + 1, 1);
  if (pos[a[x] + 1] < x && pos[a[x] + 1] >= y) add(a[x] + 1, -1);
  if (a[x] + 1 != a[y]) {
    if (y > pos[a[y] - 1] && x <= pos[a[y] - 1]) add(a[y], 1);
    if (y < pos[a[y] - 1] && x >= pos[a[y] - 1]) add(a[y], -1);
  }
  if (a[y] + 1 != a[x]) {
    if (pos[a[y] + 1] > y && pos[a[y] + 1] <= x) add(a[y] + 1, 1);
    if (pos[a[y] + 1] < y && pos[a[y] + 1] >= x) add(a[y] + 1, -1);
  }
  swap(pos[a[x]], pos[a[y]]);
  swap(a[x], a[y]);
}
int main() {
  memset(pos, 0xff, sizeof(pos));
  scanf("%d", &n);
  pos[0] = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (pos[a[i] - 1] == -1) add(a[i], 1);
    pos[a[i]] = i;
  }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int p, x, y;
    scanf("%d %d %d", &p, &x, &y);
    if (p == 1)
      query(x, y);
    else
      change(x, y);
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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 MAXN = 300005;
int n, m;
int a[MAXN], t[MAXN], pos[MAXN];
inline int lowbit(int i) { return i & (-i); }
void add(int x, int k) {
  for (int i = x; i <= n; i += lowbit(i)) t[i] += k;
}
int get(int x) {
  int res = 0;
  for (int i = x; i; i -= lowbit(i)) res += t[i];
  return res;
}
void query(int x, int y) {
  int ans = get(y) - get(x) + 1;
  printf("%d\n", ans);
}
void change(int x, int y) {
  if (x > pos[a[x] - 1] && y <= pos[a[x] - 1]) add(a[x], 1);
  if (x < pos[a[x] - 1] && y >= pos[a[x] - 1]) add(a[x], -1);
  if (pos[a[x] + 1] > x && pos[a[x] + 1] <= y) add(a[x] + 1, 1);
  if (pos[a[x] + 1] < x && pos[a[x] + 1] >= y) add(a[x] + 1, -1);
  if (a[x] + 1 != a[y]) {
    if (y > pos[a[y] - 1] && x <= pos[a[y] - 1]) add(a[y], 1);
    if (y < pos[a[y] - 1] && x >= pos[a[y] - 1]) add(a[y], -1);
  }
  if (a[y] + 1 != a[x]) {
    if (pos[a[y] + 1] > y && pos[a[y] + 1] <= x) add(a[y] + 1, 1);
    if (pos[a[y] + 1] < y && pos[a[y] + 1] >= x) add(a[y] + 1, -1);
  }
  swap(pos[a[x]], pos[a[y]]);
  swap(a[x], a[y]);
}
int main() {
  memset(pos, 0xff, sizeof(pos));
  scanf("%d", &n);
  pos[0] = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (pos[a[i] - 1] == -1) add(a[i], 1);
    pos[a[i]] = i;
  }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int p, x, y;
    scanf("%d %d %d", &p, &x, &y);
    if (p == 1)
      query(x, y);
    else
      change(x, y);
  }
  return 0;
}

```