#include <bits/stdc++.h>
using namespace std;
const int maxn = 300 * 1000 + 85 - 69;
int seg[maxn * 3], p[maxn], pos[maxn];
int n;
void modify(int idx, int val, int l = 0, int r = n, int id = 1) {
  if (idx < l or idx >= r) return;
  if (r - l == 1) {
    seg[id] = val < 0;
    return;
  }
  int mid = (l + r) >> 1;
  modify(idx, val, l, mid, id * 2 + 1);
  modify(idx, val, mid, r, id * 2 + 0);
  seg[id] = seg[id * 2 + 1] + seg[id * 2 + 0];
  return;
}
int get(int st, int en, int l = 0, int r = n, int id = 1) {
  if (st >= r or en <= l) return 0;
  if (st <= l and r <= en) return seg[id];
  int mid = (l + r) >> 1;
  return get(st, en, l, mid, id * 2 + 1) + get(st, en, mid, r, id * 2 + 0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> p[i], p[i]--;
    pos[p[i]] = i;
  }
  pos[n] = n;
  for (int i = 0; i < n; i++) modify(p[i], pos[p[i] + 1] - i);
  int q;
  cin >> q;
  while (q--) {
    int ty, fi, se;
    cin >> ty >> fi >> se;
    fi--, se--;
    if (ty == 1)
      cout << get(fi, se) + 1 << '\n';
    else {
      swap(p[fi], p[se]);
      pos[p[fi]] = fi;
      pos[p[se]] = se;
      if (p[fi]) modify(p[fi] - 1, fi - pos[p[fi] - 1]);
      modify(p[fi], pos[p[fi] + 1] - fi);
      if (p[se]) modify(p[se] - 1, se - pos[p[se] - 1]);
      modify(p[se], pos[p[se] + 1] - se);
    }
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 = 300 * 1000 + 85 - 69;
int seg[maxn * 3], p[maxn], pos[maxn];
int n;
void modify(int idx, int val, int l = 0, int r = n, int id = 1) {
  if (idx < l or idx >= r) return;
  if (r - l == 1) {
    seg[id] = val < 0;
    return;
  }
  int mid = (l + r) >> 1;
  modify(idx, val, l, mid, id * 2 + 1);
  modify(idx, val, mid, r, id * 2 + 0);
  seg[id] = seg[id * 2 + 1] + seg[id * 2 + 0];
  return;
}
int get(int st, int en, int l = 0, int r = n, int id = 1) {
  if (st >= r or en <= l) return 0;
  if (st <= l and r <= en) return seg[id];
  int mid = (l + r) >> 1;
  return get(st, en, l, mid, id * 2 + 1) + get(st, en, mid, r, id * 2 + 0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> p[i], p[i]--;
    pos[p[i]] = i;
  }
  pos[n] = n;
  for (int i = 0; i < n; i++) modify(p[i], pos[p[i] + 1] - i);
  int q;
  cin >> q;
  while (q--) {
    int ty, fi, se;
    cin >> ty >> fi >> se;
    fi--, se--;
    if (ty == 1)
      cout << get(fi, se) + 1 << '\n';
    else {
      swap(p[fi], p[se]);
      pos[p[fi]] = fi;
      pos[p[se]] = se;
      if (p[fi]) modify(p[fi] - 1, fi - pos[p[fi] - 1]);
      modify(p[fi], pos[p[fi] + 1] - fi);
      if (p[se]) modify(p[se] - 1, se - pos[p[se] - 1]);
      modify(p[se], pos[p[se] + 1] - se);
    }
  }
  return 0;
}

```