#include <bits/stdc++.h>
using namespace std;
const int maxn = 4e5;
int a[maxn], p[maxn], pre[maxn], nt[maxn], ta[maxn], n;
void add(int p, int x) {
  while (p <= n + 1) {
    ta[p] += x;
    p += p & -p;
  }
}
int sum(int x) {
  if (!x) return 0;
  int ans = 0;
  while (x) {
    ans += ta[x];
    x -= x & -x;
  }
  return ans;
}
int main() {
  int m, ord, x, y;
  scanf("%d", &n);
  p[0] = n + 1;
  p[n + 1] = 0;
  for (int i = 1; i < n + 1; i++) {
    scanf("%d", a + i);
    p[a[i]] = i;
  }
  for (int i = 1; i < n + 1; i++)
    if (p[i] < p[i - 1]) add(i, 1);
  scanf("%d", &m);
  for (int mcnt = 0; mcnt < m; mcnt++) {
    scanf("%d%d%d", &ord, &x, &y);
    if (ord == 1) {
      int ans = 0;
      ans = sum(y) - sum(x) + 1;
      printf("%d\n", ans);
    } else {
      int ax = a[x], ay = a[y], t1, t2, t3, t4, tt1, tt2, tt3, tt4;
      t1 = t2 = t3 = t4 = tt1 = tt2 = tt3 = tt4 = 0;
      if (p[ax] > p[ax - 1]) t1 = 1;
      if (p[ax + 1] > p[ax]) t2 = 1;
      if (p[ay] > p[ay - 1]) t3 = 1;
      if (p[ay + 1] > p[ay]) t4 = 1;
      int t;
      p[a[y]] = x;
      p[a[x]] = y;
      t = a[x];
      a[x] = a[y];
      a[y] = t;
      if (p[ax] > p[ax - 1]) tt1 = 1;
      if (p[ax + 1] > p[ax]) tt2 = 1;
      if (p[ay] > p[ay - 1]) tt3 = 1;
      if (p[ay + 1] > p[ay]) tt4 = 1;
      if (t1 != tt1) add(ax, t1 - tt1);
      if (ay != ax + 1 && t2 != tt2) add(ax + 1, t2 - tt2);
      if (t3 != tt3) add(ay, t3 - tt3);
      if (ax != ay + 1 && t4 != tt4) add(ay + 1, t4 - tt4);
    }
  }
  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 int maxn = 4e5;
int a[maxn], p[maxn], pre[maxn], nt[maxn], ta[maxn], n;
void add(int p, int x) {
  while (p <= n + 1) {
    ta[p] += x;
    p += p & -p;
  }
}
int sum(int x) {
  if (!x) return 0;
  int ans = 0;
  while (x) {
    ans += ta[x];
    x -= x & -x;
  }
  return ans;
}
int main() {
  int m, ord, x, y;
  scanf("%d", &n);
  p[0] = n + 1;
  p[n + 1] = 0;
  for (int i = 1; i < n + 1; i++) {
    scanf("%d", a + i);
    p[a[i]] = i;
  }
  for (int i = 1; i < n + 1; i++)
    if (p[i] < p[i - 1]) add(i, 1);
  scanf("%d", &m);
  for (int mcnt = 0; mcnt < m; mcnt++) {
    scanf("%d%d%d", &ord, &x, &y);
    if (ord == 1) {
      int ans = 0;
      ans = sum(y) - sum(x) + 1;
      printf("%d\n", ans);
    } else {
      int ax = a[x], ay = a[y], t1, t2, t3, t4, tt1, tt2, tt3, tt4;
      t1 = t2 = t3 = t4 = tt1 = tt2 = tt3 = tt4 = 0;
      if (p[ax] > p[ax - 1]) t1 = 1;
      if (p[ax + 1] > p[ax]) t2 = 1;
      if (p[ay] > p[ay - 1]) t3 = 1;
      if (p[ay + 1] > p[ay]) t4 = 1;
      int t;
      p[a[y]] = x;
      p[a[x]] = y;
      t = a[x];
      a[x] = a[y];
      a[y] = t;
      if (p[ax] > p[ax - 1]) tt1 = 1;
      if (p[ax + 1] > p[ax]) tt2 = 1;
      if (p[ay] > p[ay - 1]) tt3 = 1;
      if (p[ay + 1] > p[ay]) tt4 = 1;
      if (t1 != tt1) add(ax, t1 - tt1);
      if (ay != ax + 1 && t2 != tt2) add(ax + 1, t2 - tt2);
      if (t3 != tt3) add(ay, t3 - tt3);
      if (ax != ay + 1 && t4 != tt4) add(ay + 1, t4 - tt4);
    }
  }
  return 0;
}

```