#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 20, inf = 1e17;
int n, q, A[maxn], B[maxn], tmp1, tmp2, tmp3;
vector<int> ans;
struct node {
  int cnt, le, ri;
} seg[4 * maxn];
void build(int s, int e, int id) {
  if (e - s == 1) {
    seg[id].le = seg[id].ri = B[s];
    return;
  }
  int mid = (s + e) / 2;
  build(s, mid, id * 2);
  build(mid, e, id * 2 + 1);
  seg[id].cnt = seg[id * 2].cnt + seg[id * 2 + 1].cnt;
  if (seg[id * 2 + 1].le < seg[id * 2].ri) seg[id].cnt++;
  seg[id].le = seg[id * 2].le;
  seg[id].ri = seg[id * 2 + 1].ri;
}
void upd(int s, int e, int id, int ind) {
  if (e - s == 1) {
    seg[id].le = seg[id].ri = B[s];
    return;
  }
  int mid = (s + e) / 2;
  if (ind < mid)
    upd(s, mid, id * 2, ind);
  else
    upd(mid, e, id * 2 + 1, ind);
  seg[id].cnt = seg[id * 2].cnt + seg[id * 2 + 1].cnt;
  if (seg[id * 2 + 1].le < seg[id * 2].ri) seg[id].cnt++;
  seg[id].le = seg[id * 2].le;
  seg[id].ri = seg[id * 2 + 1].ri;
}
node get(int s, int e, int id, int l, int r) {
  node tmp, tmpp, tmppp;
  if (s >= r || e <= l) {
    tmp.cnt = 0;
    tmp.ri = -1;
    tmp.le = maxn;
    return tmp;
  }
  if (s >= l && e <= r) return seg[id];
  int mid = (s + e) / 2;
  tmpp = get(s, mid, id * 2, l, r);
  tmppp = get(mid, e, id * 2 + 1, l, r);
  tmp.cnt = tmpp.cnt + tmppp.cnt;
  if (tmppp.le < tmpp.ri) tmp.cnt++;
  tmp.le = tmpp.le;
  tmp.ri = tmppp.ri;
  return tmp;
}
int32_t main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    B[A[i]] = i;
  }
  build(1, n + 1, 1);
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> tmp1 >> tmp2 >> tmp3;
    if (tmp1 == 1) {
      node tt = get(1, n + 1, 1, tmp2, tmp3 + 1);
      ans.push_back(tt.cnt + 1);
    } else {
      swap(B[A[tmp2]], B[A[tmp3]]);
      swap(A[tmp2], A[tmp3]);
      upd(1, n + 1, 1, A[tmp2]);
      upd(1, n + 1, 1, A[tmp3]);
    }
  }
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << endl;
  }
}

### 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 int maxn = 3e5 + 20, inf = 1e17;
int n, q, A[maxn], B[maxn], tmp1, tmp2, tmp3;
vector<int> ans;
struct node {
  int cnt, le, ri;
} seg[4 * maxn];
void build(int s, int e, int id) {
  if (e - s == 1) {
    seg[id].le = seg[id].ri = B[s];
    return;
  }
  int mid = (s + e) / 2;
  build(s, mid, id * 2);
  build(mid, e, id * 2 + 1);
  seg[id].cnt = seg[id * 2].cnt + seg[id * 2 + 1].cnt;
  if (seg[id * 2 + 1].le < seg[id * 2].ri) seg[id].cnt++;
  seg[id].le = seg[id * 2].le;
  seg[id].ri = seg[id * 2 + 1].ri;
}
void upd(int s, int e, int id, int ind) {
  if (e - s == 1) {
    seg[id].le = seg[id].ri = B[s];
    return;
  }
  int mid = (s + e) / 2;
  if (ind < mid)
    upd(s, mid, id * 2, ind);
  else
    upd(mid, e, id * 2 + 1, ind);
  seg[id].cnt = seg[id * 2].cnt + seg[id * 2 + 1].cnt;
  if (seg[id * 2 + 1].le < seg[id * 2].ri) seg[id].cnt++;
  seg[id].le = seg[id * 2].le;
  seg[id].ri = seg[id * 2 + 1].ri;
}
node get(int s, int e, int id, int l, int r) {
  node tmp, tmpp, tmppp;
  if (s >= r || e <= l) {
    tmp.cnt = 0;
    tmp.ri = -1;
    tmp.le = maxn;
    return tmp;
  }
  if (s >= l && e <= r) return seg[id];
  int mid = (s + e) / 2;
  tmpp = get(s, mid, id * 2, l, r);
  tmppp = get(mid, e, id * 2 + 1, l, r);
  tmp.cnt = tmpp.cnt + tmppp.cnt;
  if (tmppp.le < tmpp.ri) tmp.cnt++;
  tmp.le = tmpp.le;
  tmp.ri = tmppp.ri;
  return tmp;
}
int32_t main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    B[A[i]] = i;
  }
  build(1, n + 1, 1);
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> tmp1 >> tmp2 >> tmp3;
    if (tmp1 == 1) {
      node tt = get(1, n + 1, 1, tmp2, tmp3 + 1);
      ans.push_back(tt.cnt + 1);
    } else {
      swap(B[A[tmp2]], B[A[tmp3]]);
      swap(A[tmp2], A[tmp3]);
      upd(1, n + 1, 1, A[tmp2]);
      upd(1, n + 1, 1, A[tmp3]);
    }
  }
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << endl;
  }
}

```