#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 100;
int tree[maxn << 2];
int a[maxn], p[maxn];
void push(int id) { tree[id] = tree[id << 1] + tree[id << 1 | 1]; }
void update(int ind, int x, int l, int r, int id) {
  if (l == r) {
    tree[id] = x;
    return;
  }
  int mid = (l + r) >> 1;
  if (ind <= mid)
    update(ind, x, l, mid, id << 1);
  else
    update(ind, x, mid + 1, r, id << 1 | 1);
  push(id);
}
int query(int L, int R, int l, int r, int id) {
  if (L <= l && r <= R) return tree[id];
  int mid = (l + r) >> 1;
  int res = 0;
  if (L <= mid) res += query(L, R, l, mid, id << 1);
  if (mid < R) res += query(L, R, mid + 1, r, id << 1 | 1);
  return res;
}
int main() {
  int n, q;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), p[a[i]] = i;
  p[0] = 0, p[n + 1] = n + 1;
  for (int i = 2; i <= n; i++)
    if (p[i] < p[i - 1]) update(i, 1, 1, n, 1);
  scanf("%d", &q);
  int op, x, y;
  while (q--) {
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1)
      printf("%d\n", query(x + 1, y, 1, n, 1) + 1);
    else {
      int xx = a[x], yy = a[y];
      swap(a[x], a[y]);
      swap(p[xx], p[yy]);
      update(xx, 0, 1, n, 1);
      update(xx + 1, 0, 1, n, 1);
      update(yy, 0, 1, n, 1);
      update(yy + 1, 0, 1, n, 1);
      if (p[xx] < p[xx - 1]) update(xx, 1, 1, n, 1);
      if (p[xx + 1] < p[xx]) update(xx + 1, 1, 1, n, 1);
      if (p[yy] < p[yy - 1]) update(yy, 1, 1, n, 1);
      if (p[yy + 1] < p[yy]) update(yy + 1, 1, 1, n, 1);
    }
  }
  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 = 3e5 + 100;
int tree[maxn << 2];
int a[maxn], p[maxn];
void push(int id) { tree[id] = tree[id << 1] + tree[id << 1 | 1]; }
void update(int ind, int x, int l, int r, int id) {
  if (l == r) {
    tree[id] = x;
    return;
  }
  int mid = (l + r) >> 1;
  if (ind <= mid)
    update(ind, x, l, mid, id << 1);
  else
    update(ind, x, mid + 1, r, id << 1 | 1);
  push(id);
}
int query(int L, int R, int l, int r, int id) {
  if (L <= l && r <= R) return tree[id];
  int mid = (l + r) >> 1;
  int res = 0;
  if (L <= mid) res += query(L, R, l, mid, id << 1);
  if (mid < R) res += query(L, R, mid + 1, r, id << 1 | 1);
  return res;
}
int main() {
  int n, q;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), p[a[i]] = i;
  p[0] = 0, p[n + 1] = n + 1;
  for (int i = 2; i <= n; i++)
    if (p[i] < p[i - 1]) update(i, 1, 1, n, 1);
  scanf("%d", &q);
  int op, x, y;
  while (q--) {
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1)
      printf("%d\n", query(x + 1, y, 1, n, 1) + 1);
    else {
      int xx = a[x], yy = a[y];
      swap(a[x], a[y]);
      swap(p[xx], p[yy]);
      update(xx, 0, 1, n, 1);
      update(xx + 1, 0, 1, n, 1);
      update(yy, 0, 1, n, 1);
      update(yy + 1, 0, 1, n, 1);
      if (p[xx] < p[xx - 1]) update(xx, 1, 1, n, 1);
      if (p[xx + 1] < p[xx]) update(xx + 1, 1, 1, n, 1);
      if (p[yy] < p[yy - 1]) update(yy, 1, 1, n, 1);
      if (p[yy + 1] < p[yy]) update(yy + 1, 1, 1, n, 1);
    }
  }
  return 0;
}

```