#include <bits/stdc++.h>
using namespace std;
struct node {
  int ans, l, r;
};
int n, m, a[((int)301 * 1000)], pos[((int)301 * 1000)];
node seg[4 * ((int)301 * 1000)];
node merge(node x, node y) {
  node res;
  res.l = x.l;
  res.r = y.r;
  res.ans = x.ans + y.ans + (x.r > y.l);
  return res;
}
void update(int q, int xl = 0, int xr = ((int)301 * 1000), int id = 1) {
  if (xl == xr - 1) {
    seg[id].l = seg[id].r = pos[xl];
    return;
  }
  int mid = (xl + xr) / 2;
  if (q < mid)
    update(q, xl, mid, id * 2);
  else
    update(q, mid, xr, id * 2 + 1);
  seg[id] = merge(seg[id * 2], seg[id * 2 + 1]);
}
node query(int ql, int qr, int xl = 0, int xr = ((int)301 * 1000), int id = 1) {
  if (qr <= xl || xr <= ql) return {0, ((int)301 * 1000), -1};
  if (ql <= xl && xr <= qr) return seg[id];
  int mid = (xl + xr) / 2;
  return merge(query(ql, qr, xl, mid, id * 2),
               query(ql, qr, mid, xr, id * 2 + 1));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pos[a[i]] = i;
    update(a[i]);
  }
  cin >> m;
  while (m--) {
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) cout << query(x, y + 1).ans + 1 << "\n";
    if (t == 2)
      swap(a[x], a[y]), swap(pos[a[x]], pos[a[y]]), update(a[x]), update(a[y]);
  }
  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;
struct node {
  int ans, l, r;
};
int n, m, a[((int)301 * 1000)], pos[((int)301 * 1000)];
node seg[4 * ((int)301 * 1000)];
node merge(node x, node y) {
  node res;
  res.l = x.l;
  res.r = y.r;
  res.ans = x.ans + y.ans + (x.r > y.l);
  return res;
}
void update(int q, int xl = 0, int xr = ((int)301 * 1000), int id = 1) {
  if (xl == xr - 1) {
    seg[id].l = seg[id].r = pos[xl];
    return;
  }
  int mid = (xl + xr) / 2;
  if (q < mid)
    update(q, xl, mid, id * 2);
  else
    update(q, mid, xr, id * 2 + 1);
  seg[id] = merge(seg[id * 2], seg[id * 2 + 1]);
}
node query(int ql, int qr, int xl = 0, int xr = ((int)301 * 1000), int id = 1) {
  if (qr <= xl || xr <= ql) return {0, ((int)301 * 1000), -1};
  if (ql <= xl && xr <= qr) return seg[id];
  int mid = (xl + xr) / 2;
  return merge(query(ql, qr, xl, mid, id * 2),
               query(ql, qr, mid, xr, id * 2 + 1));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pos[a[i]] = i;
    update(a[i]);
  }
  cin >> m;
  while (m--) {
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) cout << query(x, y + 1).ans + 1 << "\n";
    if (t == 2)
      swap(a[x], a[y]), swap(pos[a[x]], pos[a[y]]), update(a[x]), update(a[y]);
  }
  return 0;
}

```