#include <bits/stdc++.h>
inline int in() {
  int x;
  scanf("%d", &x);
  return x;
}
using namespace std;
constexpr int mod = 1e9 + 7, N = 3e5 + 7;
int a[N], ps[N], n;
struct node {
  int l, r, ans;
} seg[4 * N];
node make(node x, node y) {
  node ans;
  ans.l = x.l, ans.r = y.r, ans.ans = x.ans + y.ans;
  if (x.r < y.l) ans.ans--;
  return ans;
}
int pos[N];
int pis[N];
void build(int tag = 1, int s = 0, int e = n) {
  if (e - s == 1) {
    seg[tag].l = seg[tag].r = pis[s], seg[tag].ans = 1;
    return;
  }
  int mid = (s + e) / 2, L = tag * 2, R = tag * 2 + 1;
  build(L, s, mid);
  build(R, mid, e);
  seg[tag] = make(seg[L], seg[R]);
}
node query(int l, int r, int tag = 1, int s = 0, int e = n) {
  int mid = (s + e) / 2, L = tag * 2, R = tag * 2 + 1;
  bool b1 = !(s >= r || mid <= l);
  bool b2 = !(mid >= r || e <= l);
  if (s >= l && e <= r) {
    return seg[tag];
  }
  if (b1 && b2)
    return make(query(l, r, L, s, mid), query(l, r, R, mid, e));
  else if (b1)
    return query(l, r, L, s, mid);
  else
    return query(l, r, R, mid, e);
}
void update(int idx, int nw, int tag = 1, int s = 0, int e = n) {
  if (e - s == 1) {
    seg[tag].l = seg[tag].r = nw, seg[tag].ans = 1;
    return;
  }
  int L = tag * 2, R = tag * 2 + 1, mid = (s + e) / 2;
  if (idx < mid)
    update(idx, nw, L, s, mid);
  else
    update(idx, nw, R, mid, e);
  seg[tag] = make(seg[L], seg[R]);
}
int32_t main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x = in() - 1;
    pos[i] = x;
    pis[x] = i;
  }
  build();
  int Q = in();
  while (Q--) {
    if (in() == 1) {
      int l = in() - 1, r = in();
      cout << query(l, r).ans << endl;
    } else {
      int x = in() - 1, y = in() - 1;
      update(pos[x], y);
      update(pos[y], x);
      swap(pos[x], pos[y]);
    }
  }
}

### Prompt

In cpp, your task is to solve 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>
inline int in() {
  int x;
  scanf("%d", &x);
  return x;
}
using namespace std;
constexpr int mod = 1e9 + 7, N = 3e5 + 7;
int a[N], ps[N], n;
struct node {
  int l, r, ans;
} seg[4 * N];
node make(node x, node y) {
  node ans;
  ans.l = x.l, ans.r = y.r, ans.ans = x.ans + y.ans;
  if (x.r < y.l) ans.ans--;
  return ans;
}
int pos[N];
int pis[N];
void build(int tag = 1, int s = 0, int e = n) {
  if (e - s == 1) {
    seg[tag].l = seg[tag].r = pis[s], seg[tag].ans = 1;
    return;
  }
  int mid = (s + e) / 2, L = tag * 2, R = tag * 2 + 1;
  build(L, s, mid);
  build(R, mid, e);
  seg[tag] = make(seg[L], seg[R]);
}
node query(int l, int r, int tag = 1, int s = 0, int e = n) {
  int mid = (s + e) / 2, L = tag * 2, R = tag * 2 + 1;
  bool b1 = !(s >= r || mid <= l);
  bool b2 = !(mid >= r || e <= l);
  if (s >= l && e <= r) {
    return seg[tag];
  }
  if (b1 && b2)
    return make(query(l, r, L, s, mid), query(l, r, R, mid, e));
  else if (b1)
    return query(l, r, L, s, mid);
  else
    return query(l, r, R, mid, e);
}
void update(int idx, int nw, int tag = 1, int s = 0, int e = n) {
  if (e - s == 1) {
    seg[tag].l = seg[tag].r = nw, seg[tag].ans = 1;
    return;
  }
  int L = tag * 2, R = tag * 2 + 1, mid = (s + e) / 2;
  if (idx < mid)
    update(idx, nw, L, s, mid);
  else
    update(idx, nw, R, mid, e);
  seg[tag] = make(seg[L], seg[R]);
}
int32_t main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x = in() - 1;
    pos[i] = x;
    pis[x] = i;
  }
  build();
  int Q = in();
  while (Q--) {
    if (in() == 1) {
      int l = in() - 1, r = in();
      cout << query(l, r).ans << endl;
    } else {
      int x = in() - 1, y = in() - 1;
      update(pos[x], y);
      update(pos[y], x);
      swap(pos[x], pos[y]);
    }
  }
}

```