#include <bits/stdc++.h>
using namespace std;
int tmp[311111], position[311111], segm[4 * 311111];
int n;
int func(int md) { return position[md - 1] > position[md]; }
int get_ans(int v, int l, int r, int tl, int tr) {
  if (l == tl && r == tr) return segm[v];
  int md = (l + r) >> 1;
  if (tr <= md)
    return get_ans(2 * v, l, md, tl, tr);
  else if (tl > md)
    return get_ans(2 * v + 1, md + 1, r, tl, tr);
  return get_ans(2 * v, l, md, tl, md) +
         get_ans(2 * v + 1, md + 1, r, md + 1, tr) + func(md);
}
void uddate(int v, int l, int r, int p) {
  if (l == r)
    segm[v] = 0;
  else {
    int md = (l + r) >> 1;
    if (p <= md)
      uddate(2 * v, l, md, p);
    else
      uddate(2 * v + 1, md + 1, r, p);
    segm[v] = segm[2 * v] + segm[2 * v + 1] + func(md);
  }
}
int get_ans(int l, int r) { return get_ans(1, 1, n, l + 1, r + 1); }
void uddate(int p) {
  if (p < 0 || p > n - 1) return;
  uddate(1, 1, n, p + 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> tmp[i];
    position[--tmp[i]] = i;
  }
  for (int i = 0; i < n; ++i) uddate(i);
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int t, x, y;
    cin >> t >> x >> y;
    --x, --y;
    if (t == 1)
      cout << get_ans(x, y) + 1 << "\n";
    else {
      swap(position[tmp[x]], position[tmp[y]]);
      swap(tmp[x], tmp[y]);
      uddate(tmp[x]);
      uddate(tmp[y]);
      uddate(tmp[x] - 1);
      uddate(tmp[y] + 1);
      uddate(tmp[x] + 1);
      uddate(tmp[y] + 1);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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 tmp[311111], position[311111], segm[4 * 311111];
int n;
int func(int md) { return position[md - 1] > position[md]; }
int get_ans(int v, int l, int r, int tl, int tr) {
  if (l == tl && r == tr) return segm[v];
  int md = (l + r) >> 1;
  if (tr <= md)
    return get_ans(2 * v, l, md, tl, tr);
  else if (tl > md)
    return get_ans(2 * v + 1, md + 1, r, tl, tr);
  return get_ans(2 * v, l, md, tl, md) +
         get_ans(2 * v + 1, md + 1, r, md + 1, tr) + func(md);
}
void uddate(int v, int l, int r, int p) {
  if (l == r)
    segm[v] = 0;
  else {
    int md = (l + r) >> 1;
    if (p <= md)
      uddate(2 * v, l, md, p);
    else
      uddate(2 * v + 1, md + 1, r, p);
    segm[v] = segm[2 * v] + segm[2 * v + 1] + func(md);
  }
}
int get_ans(int l, int r) { return get_ans(1, 1, n, l + 1, r + 1); }
void uddate(int p) {
  if (p < 0 || p > n - 1) return;
  uddate(1, 1, n, p + 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> tmp[i];
    position[--tmp[i]] = i;
  }
  for (int i = 0; i < n; ++i) uddate(i);
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int t, x, y;
    cin >> t >> x >> y;
    --x, --y;
    if (t == 1)
      cout << get_ans(x, y) + 1 << "\n";
    else {
      swap(position[tmp[x]], position[tmp[y]]);
      swap(tmp[x], tmp[y]);
      uddate(tmp[x]);
      uddate(tmp[y]);
      uddate(tmp[x] - 1);
      uddate(tmp[y] + 1);
      uddate(tmp[x] + 1);
      uddate(tmp[y] + 1);
    }
  }
  return 0;
}

```