#include <bits/stdc++.h>
using namespace std;
const long long N = 4e6 + 100, OO = 1e6 + 10, M = 1e9 + 7, lg = 22;
long long a[N], b[N], seg[N];
void build(long long c, long long b, long long e) {
  if (e - b == 1) {
    seg[c] = 1;
    return;
  }
  long long mid = (b + e) / 2;
  build(2 * c, b, mid);
  build(2 * c + 1, mid, e);
  seg[c] = seg[2 * c] + seg[2 * c + 1] - (a[mid - 1] < a[mid]);
}
long long get(long long c, long long b, long long e, long long l, long long r) {
  if (b == l && e == r) return seg[c];
  long long mid = (b + e) / 2;
  if (r <= mid) return get(2 * c, b, mid, l, r);
  if (l >= mid) return get(2 * c + 1, mid, e, l, r);
  return get(2 * c, b, mid, l, mid) + get(2 * c + 1, mid, e, mid, r) -
         (a[mid - 1] < a[mid]);
}
void upd(long long c, long long b, long long e, long long x) {
  if (e - b == 1) {
    seg[c] = 1;
    return;
  }
  long long mid = (b + e) / 2;
  if (x < mid)
    upd(2 * c, b, mid, x);
  else
    upd(2 * c + 1, mid, e, x);
  seg[c] = seg[2 * c] + seg[2 * c + 1] - (a[mid - 1] < a[mid]);
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> b[i];
    b[i]--;
    a[b[i]] = i;
  }
  vector<pair<long long, long long> > v;
  build(1, 0, n);
  long long q;
  cin >> q;
  while (q--) {
    long long t, x, y;
    cin >> t >> x >> y;
    x--;
    if (t == 1) {
      cout << get(1, 0, n, x, y) << endl;
      continue;
    } else {
      v.push_back({x + 1, y});
    }
    y--;
    swap(b[x], b[y]);
    a[b[x]] = x;
    a[b[y]] = y;
    upd(1, 0, n, b[x]);
    upd(1, 0, n, b[y]);
  }
  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;
const long long N = 4e6 + 100, OO = 1e6 + 10, M = 1e9 + 7, lg = 22;
long long a[N], b[N], seg[N];
void build(long long c, long long b, long long e) {
  if (e - b == 1) {
    seg[c] = 1;
    return;
  }
  long long mid = (b + e) / 2;
  build(2 * c, b, mid);
  build(2 * c + 1, mid, e);
  seg[c] = seg[2 * c] + seg[2 * c + 1] - (a[mid - 1] < a[mid]);
}
long long get(long long c, long long b, long long e, long long l, long long r) {
  if (b == l && e == r) return seg[c];
  long long mid = (b + e) / 2;
  if (r <= mid) return get(2 * c, b, mid, l, r);
  if (l >= mid) return get(2 * c + 1, mid, e, l, r);
  return get(2 * c, b, mid, l, mid) + get(2 * c + 1, mid, e, mid, r) -
         (a[mid - 1] < a[mid]);
}
void upd(long long c, long long b, long long e, long long x) {
  if (e - b == 1) {
    seg[c] = 1;
    return;
  }
  long long mid = (b + e) / 2;
  if (x < mid)
    upd(2 * c, b, mid, x);
  else
    upd(2 * c + 1, mid, e, x);
  seg[c] = seg[2 * c] + seg[2 * c + 1] - (a[mid - 1] < a[mid]);
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> b[i];
    b[i]--;
    a[b[i]] = i;
  }
  vector<pair<long long, long long> > v;
  build(1, 0, n);
  long long q;
  cin >> q;
  while (q--) {
    long long t, x, y;
    cin >> t >> x >> y;
    x--;
    if (t == 1) {
      cout << get(1, 0, n, x, y) << endl;
      continue;
    } else {
      v.push_back({x + 1, y});
    }
    y--;
    swap(b[x], b[y]);
    a[b[x]] = x;
    a[b[y]] = y;
    upd(1, 0, n, b[x]);
    upd(1, 0, n, b[y]);
  }
  return 0;
}

```