#include <bits/stdc++.h>
using namespace std;
const int MAXN = 300000;
int a[MAXN], pos[MAXN];
int tree[MAXN * 4];
void update(int x, int l, int r, int p, int v) {
  if (l == r) {
    tree[x] += v;
    return;
  }
  int mid = (l + r) / 2;
  if (p <= mid)
    update(x * 2, l, mid, p, v);
  else
    update(x * 2 + 1, mid + 1, r, p, v);
  tree[x] = tree[x * 2] + tree[x * 2 + 1];
}
int query(int x, int l, int r, int s, int t) {
  if (s <= l && r <= t) return tree[x];
  int mid = (l + r) / 2;
  int ret = 0;
  if (s <= mid) ret = query(x * 2, l, mid, s, t);
  if (t > mid) ret += query(x * 2 + 1, mid + 1, r, s, t);
  return ret;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) a[i] = i + 1;
  for (int i = 0; i < 6; ++i) next_permutation(a, a + n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    pos[--a[i]] = i;
  }
  for (int i = 0; i + 1 < n; ++i)
    if (pos[i] > pos[i + 1]) update(1, 0, n - 1, i, 1);
  int m;
  scanf("%d", &m);
  while (m--) {
    int typ, l, r;
    scanf("%d%d%d", &typ, &l, &r);
    if (typ == 1) {
      printf("%d\n", 1 + query(1, 0, n - 1, l - 1, r - 1 - 1));
    } else {
      int x = a[--l], y = a[--r];
      if (x > y) swap(x, y);
      {
        if (x) {
          if (pos[x - 1] > pos[x]) update(1, 0, n - 1, x - 1, -1);
          if (pos[x - 1] > pos[y]) update(1, 0, n - 1, x - 1, 1);
        }
        if (x + 1 < y) {
          if (pos[x + 1] < pos[x]) update(1, 0, n - 1, x, -1);
          if (pos[x + 1] < pos[y]) update(1, 0, n - 1, x, 1);
          if (pos[y - 1] > pos[y]) update(1, 0, n - 1, y - 1, -1);
          if (pos[y - 1] > pos[x]) update(1, 0, n - 1, y - 1, 1);
        } else {
          if (pos[x] > pos[y])
            update(1, 0, n - 1, x, -1);
          else
            update(1, 0, n - 1, x, 1);
        }
        if (y + 1 < n) {
          if (pos[y + 1] < pos[y]) update(1, 0, n - 1, y, -1);
          if (pos[y + 1] < pos[x]) update(1, 0, n - 1, y, 1);
        }
      }
      swap(a[l], a[r]);
      pos[a[l]] = l;
      pos[a[r]] = r;
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
const int MAXN = 300000;
int a[MAXN], pos[MAXN];
int tree[MAXN * 4];
void update(int x, int l, int r, int p, int v) {
  if (l == r) {
    tree[x] += v;
    return;
  }
  int mid = (l + r) / 2;
  if (p <= mid)
    update(x * 2, l, mid, p, v);
  else
    update(x * 2 + 1, mid + 1, r, p, v);
  tree[x] = tree[x * 2] + tree[x * 2 + 1];
}
int query(int x, int l, int r, int s, int t) {
  if (s <= l && r <= t) return tree[x];
  int mid = (l + r) / 2;
  int ret = 0;
  if (s <= mid) ret = query(x * 2, l, mid, s, t);
  if (t > mid) ret += query(x * 2 + 1, mid + 1, r, s, t);
  return ret;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) a[i] = i + 1;
  for (int i = 0; i < 6; ++i) next_permutation(a, a + n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    pos[--a[i]] = i;
  }
  for (int i = 0; i + 1 < n; ++i)
    if (pos[i] > pos[i + 1]) update(1, 0, n - 1, i, 1);
  int m;
  scanf("%d", &m);
  while (m--) {
    int typ, l, r;
    scanf("%d%d%d", &typ, &l, &r);
    if (typ == 1) {
      printf("%d\n", 1 + query(1, 0, n - 1, l - 1, r - 1 - 1));
    } else {
      int x = a[--l], y = a[--r];
      if (x > y) swap(x, y);
      {
        if (x) {
          if (pos[x - 1] > pos[x]) update(1, 0, n - 1, x - 1, -1);
          if (pos[x - 1] > pos[y]) update(1, 0, n - 1, x - 1, 1);
        }
        if (x + 1 < y) {
          if (pos[x + 1] < pos[x]) update(1, 0, n - 1, x, -1);
          if (pos[x + 1] < pos[y]) update(1, 0, n - 1, x, 1);
          if (pos[y - 1] > pos[y]) update(1, 0, n - 1, y - 1, -1);
          if (pos[y - 1] > pos[x]) update(1, 0, n - 1, y - 1, 1);
        } else {
          if (pos[x] > pos[y])
            update(1, 0, n - 1, x, -1);
          else
            update(1, 0, n - 1, x, 1);
        }
        if (y + 1 < n) {
          if (pos[y + 1] < pos[y]) update(1, 0, n - 1, y, -1);
          if (pos[y + 1] < pos[x]) update(1, 0, n - 1, y, 1);
        }
      }
      swap(a[l], a[r]);
      pos[a[l]] = l;
      pos[a[r]] = r;
    }
  }
  return 0;
}

```