#include <bits/stdc++.h>
using namespace std;
int a[350001], p[350001], t[1500001], n, i, q, type, l, r, nom1, nom2;
void build(int l, int r, int nom) {
  if (l == r) {
    t[nom] = (p[l] > p[l + 1]);
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, nom * 2);
  build(mid + 1, r, nom * 2 + 1);
  t[nom] = t[nom * 2] + t[nom * 2 + 1];
}
int get(int l, int r, int nom, int l1, int r1) {
  if (r < l1 || l > r1) return 0;
  if (l1 <= l && r <= r1) return t[nom];
  int mid = (l + r) / 2;
  return get(l, mid, nom * 2, l1, r1) + get(mid + 1, r, nom * 2 + 1, l1, r1);
}
void update(int l, int r, int nom, int nom1, int val) {
  if (l > nom1 || r < nom1) return;
  t[nom] += val;
  if (l == r) return;
  int mid = (l + r) / 2;
  update(l, mid, nom * 2, nom1, val);
  update(mid + 1, r, nom * 2 + 1, nom1, val);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    p[a[i]] = i;
  }
  build(1, n - 1, 1);
  cin >> q;
  while (q--) {
    cin >> type;
    if (type == 1) {
      cin >> l >> r;
      cout << get(1, n - 1, 1, l, r - 1) + 1 << '\n';
    } else {
      cin >> nom1 >> nom2;
      swap(a[nom1], a[nom2]);
      nom1 = a[nom1];
      nom2 = a[nom2];
      if (nom1 > nom2) swap(nom1, nom2);
      if (nom1 < n) {
        if (p[nom1] > p[nom1 + 1]) update(1, n - 1, 1, nom1, -1);
      }
      if (nom1 - 1 >= 1) {
        if (p[nom1 - 1] > p[nom1]) update(1, n - 1, 1, nom1 - 1, -1);
      }
      if (nom2 < n) {
        if (p[nom2] > p[nom2 + 1]) update(1, n - 1, 1, nom2, -1);
      }
      if (nom2 - 1 >= 1 && nom2 - 1 != nom1) {
        if (p[nom2 - 1] > p[nom2]) update(1, n - 1, 1, nom2 - 1, -1);
      }
      swap(p[nom1], p[nom2]);
      if (nom1 < n) {
        if (p[nom1] > p[nom1 + 1]) update(1, n - 1, 1, nom1, 1);
      }
      if (nom1 - 1 >= 1) {
        if (p[nom1 - 1] > p[nom1]) update(1, n - 1, 1, nom1 - 1, 1);
      }
      if (nom2 < n) {
        if (p[nom2] > p[nom2 + 1]) update(1, n - 1, 1, nom2, 1);
      }
      if (nom2 - 1 >= 1 && nom2 - 1 != nom1) {
        if (p[nom2 - 1] > p[nom2]) update(1, n - 1, 1, nom2 - 1, 1);
      }
    }
  }
  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;
int a[350001], p[350001], t[1500001], n, i, q, type, l, r, nom1, nom2;
void build(int l, int r, int nom) {
  if (l == r) {
    t[nom] = (p[l] > p[l + 1]);
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, nom * 2);
  build(mid + 1, r, nom * 2 + 1);
  t[nom] = t[nom * 2] + t[nom * 2 + 1];
}
int get(int l, int r, int nom, int l1, int r1) {
  if (r < l1 || l > r1) return 0;
  if (l1 <= l && r <= r1) return t[nom];
  int mid = (l + r) / 2;
  return get(l, mid, nom * 2, l1, r1) + get(mid + 1, r, nom * 2 + 1, l1, r1);
}
void update(int l, int r, int nom, int nom1, int val) {
  if (l > nom1 || r < nom1) return;
  t[nom] += val;
  if (l == r) return;
  int mid = (l + r) / 2;
  update(l, mid, nom * 2, nom1, val);
  update(mid + 1, r, nom * 2 + 1, nom1, val);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    p[a[i]] = i;
  }
  build(1, n - 1, 1);
  cin >> q;
  while (q--) {
    cin >> type;
    if (type == 1) {
      cin >> l >> r;
      cout << get(1, n - 1, 1, l, r - 1) + 1 << '\n';
    } else {
      cin >> nom1 >> nom2;
      swap(a[nom1], a[nom2]);
      nom1 = a[nom1];
      nom2 = a[nom2];
      if (nom1 > nom2) swap(nom1, nom2);
      if (nom1 < n) {
        if (p[nom1] > p[nom1 + 1]) update(1, n - 1, 1, nom1, -1);
      }
      if (nom1 - 1 >= 1) {
        if (p[nom1 - 1] > p[nom1]) update(1, n - 1, 1, nom1 - 1, -1);
      }
      if (nom2 < n) {
        if (p[nom2] > p[nom2 + 1]) update(1, n - 1, 1, nom2, -1);
      }
      if (nom2 - 1 >= 1 && nom2 - 1 != nom1) {
        if (p[nom2 - 1] > p[nom2]) update(1, n - 1, 1, nom2 - 1, -1);
      }
      swap(p[nom1], p[nom2]);
      if (nom1 < n) {
        if (p[nom1] > p[nom1 + 1]) update(1, n - 1, 1, nom1, 1);
      }
      if (nom1 - 1 >= 1) {
        if (p[nom1 - 1] > p[nom1]) update(1, n - 1, 1, nom1 - 1, 1);
      }
      if (nom2 < n) {
        if (p[nom2] > p[nom2 + 1]) update(1, n - 1, 1, nom2, 1);
      }
      if (nom2 - 1 >= 1 && nom2 - 1 != nom1) {
        if (p[nom2 - 1] > p[nom2]) update(1, n - 1, 1, nom2 - 1, 1);
      }
    }
  }
  return 0;
}

```