#include <bits/stdc++.h>
using namespace std;
int n;
int tree[300010 + 1];
int a[300010 + 1];
int get(int pos) {
  int ret = 0, i;
  for (i = pos; i; i -= (i & -i)) ret += tree[i];
  return ret;
}
void update(int pos, int val) {
  int i;
  for (i = pos; i <= 300010; i += (i & -i)) tree[i] += val;
}
int vis[300010];
void rem(int val) {
  int l, r;
  if (get(val) - get(val - 1)) {
    update(val, -1);
  }
  if (val != n && (get(val + 1) - get(val)) == 0) {
    update(val + 1, 1);
  }
}
void add(int val) {
  if (val != n) {
    if (vis[val + 1] > vis[val])
      if (get(val + 1) - get(val)) {
        update(val + 1, -1);
      }
  }
  update(val, 1);
}
void chek(int val) {
  if (val != 1) {
    if (vis[val - 1] < vis[val]) {
      if (get(val) - get(val - 1)) update(val, -1);
    }
  }
}
int main() {
  int i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d\n", &a[i]);
    vis[a[i]] = i;
  }
  vis[0] = 0;
  vis[n + 1] = 0;
  int now;
  for (i = 1; i <= n;) {
    now = i;
    update(i, 1);
    while (vis[now + 1] > vis[now]) now++;
    i = now + 1;
  }
  int q, x, y;
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &i);
    if (i == 1) {
      scanf("%d %d", &x, &y);
      int ret = get(y) - get(x - 1);
      printf("%d\n", get(y) - get(x) + 1);
    } else {
      scanf("%d %d", &x, &y);
      swap(a[x], a[y]);
      x = a[x];
      y = a[y];
      if (x > y) swap(x, y);
      rem(x);
      rem(y);
      swap(vis[x], vis[y]);
      add(y);
      add(x);
      chek(y);
      chek(x);
    }
  }
  return 0;
}

### Prompt

Your task is to create 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;
int n;
int tree[300010 + 1];
int a[300010 + 1];
int get(int pos) {
  int ret = 0, i;
  for (i = pos; i; i -= (i & -i)) ret += tree[i];
  return ret;
}
void update(int pos, int val) {
  int i;
  for (i = pos; i <= 300010; i += (i & -i)) tree[i] += val;
}
int vis[300010];
void rem(int val) {
  int l, r;
  if (get(val) - get(val - 1)) {
    update(val, -1);
  }
  if (val != n && (get(val + 1) - get(val)) == 0) {
    update(val + 1, 1);
  }
}
void add(int val) {
  if (val != n) {
    if (vis[val + 1] > vis[val])
      if (get(val + 1) - get(val)) {
        update(val + 1, -1);
      }
  }
  update(val, 1);
}
void chek(int val) {
  if (val != 1) {
    if (vis[val - 1] < vis[val]) {
      if (get(val) - get(val - 1)) update(val, -1);
    }
  }
}
int main() {
  int i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d\n", &a[i]);
    vis[a[i]] = i;
  }
  vis[0] = 0;
  vis[n + 1] = 0;
  int now;
  for (i = 1; i <= n;) {
    now = i;
    update(i, 1);
    while (vis[now + 1] > vis[now]) now++;
    i = now + 1;
  }
  int q, x, y;
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &i);
    if (i == 1) {
      scanf("%d %d", &x, &y);
      int ret = get(y) - get(x - 1);
      printf("%d\n", get(y) - get(x) + 1);
    } else {
      scanf("%d %d", &x, &y);
      swap(a[x], a[y]);
      x = a[x];
      y = a[y];
      if (x > y) swap(x, y);
      rem(x);
      rem(y);
      swap(vis[x], vis[y]);
      add(y);
      add(x);
      chek(y);
      chek(x);
    }
  }
  return 0;
}

```