#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long maxn = 4e5;
const long long mod = 1e9 + 7;
const long double PI = acos((long double)-1);
long long pw(long long a, long long b, long long md = mod) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (a * res) % md;
    }
    a = (a * a) % md;
    b >>= 1;
  }
  return (res);
}
int n;
int q;
int a[maxn];
int pos[maxn];
int fen[maxn];
void add(int x, int pos) {
  for (pos += 5; pos < maxn; pos += pos & -pos) fen[pos] += x;
}
int get(int pos) {
  int ans = 0;
  for (pos += 5; pos; pos -= pos & -pos) ans += fen[pos];
  return (ans);
}
int get(int l, int r) { return (get(r) - get(l - 1)); }
void rem(int x) {
  if (x > 1)
    if (pos[x] < pos[x - 1] and get(x - 1, x - 1)) add(-1, x - 1);
  if (x < n)
    if (pos[x] > pos[x + 1] and get(x, x)) add(-1, x);
}
void Add(int x) {
  if (x > 1)
    if (pos[x] < pos[x - 1] and !get(x - 1, x - 1)) add(1, x - 1);
  if (x < n)
    if (pos[x] > pos[x + 1] and !get(x, x)) add(1, x);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], pos[a[i]] = i;
  for (int i = 1; i < n; i++)
    if (pos[i] > pos[i + 1]) add(1, i);
  cin >> q;
  while (q--) {
    int t, l, r;
    cin >> t >> l >> r;
    if (t == 1)
      cout << get(l, r - 1) + 1 << '\n';
    else {
      rem(a[l]);
      rem(a[r]);
      swap(pos[a[l]], pos[a[r]]);
      swap(a[l], a[r]);
      Add(a[l]);
      Add(a[r]);
    }
  }
  return (0);
}

### 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>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long maxn = 4e5;
const long long mod = 1e9 + 7;
const long double PI = acos((long double)-1);
long long pw(long long a, long long b, long long md = mod) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (a * res) % md;
    }
    a = (a * a) % md;
    b >>= 1;
  }
  return (res);
}
int n;
int q;
int a[maxn];
int pos[maxn];
int fen[maxn];
void add(int x, int pos) {
  for (pos += 5; pos < maxn; pos += pos & -pos) fen[pos] += x;
}
int get(int pos) {
  int ans = 0;
  for (pos += 5; pos; pos -= pos & -pos) ans += fen[pos];
  return (ans);
}
int get(int l, int r) { return (get(r) - get(l - 1)); }
void rem(int x) {
  if (x > 1)
    if (pos[x] < pos[x - 1] and get(x - 1, x - 1)) add(-1, x - 1);
  if (x < n)
    if (pos[x] > pos[x + 1] and get(x, x)) add(-1, x);
}
void Add(int x) {
  if (x > 1)
    if (pos[x] < pos[x - 1] and !get(x - 1, x - 1)) add(1, x - 1);
  if (x < n)
    if (pos[x] > pos[x + 1] and !get(x, x)) add(1, x);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], pos[a[i]] = i;
  for (int i = 1; i < n; i++)
    if (pos[i] > pos[i + 1]) add(1, i);
  cin >> q;
  while (q--) {
    int t, l, r;
    cin >> t >> l >> r;
    if (t == 1)
      cout << get(l, r - 1) + 1 << '\n';
    else {
      rem(a[l]);
      rem(a[r]);
      swap(pos[a[l]], pos[a[r]]);
      swap(a[l], a[r]);
      Add(a[l]);
      Add(a[r]);
    }
  }
  return (0);
}

```