#include <bits/stdc++.h>
using namespace std;
int const N = 3e5 + 20;
int n, q, a[N], p[N];
int seg[N << 2];
void modify(int idx, int val, int v = 1, int s = 0, int e = n) {
  if (s + 1 == e) {
    seg[v] = val;
    return;
  }
  int mid = (s + e) >> 1, lc = v << 1, rc = lc | 1;
  if (idx < mid)
    modify(idx, val, lc, s, mid);
  else
    modify(idx, val, rc, mid, e);
  seg[v] = seg[lc] + seg[rc];
}
int get(int l, int r, int v = 1, int s = 0, int e = n) {
  if (s >= r || e <= l) return 0;
  if (s >= l && e <= r) return seg[v];
  int mid = (s + e) >> 1, lc = v << 1, rc = lc | 1;
  return get(l, r, lc, s, mid) + get(l, r, rc, mid, e);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], a[i]--, p[a[i]] = i;
  for (int i = 0; i < n - 1; i++)
    if (p[i] > p[i + 1]) modify(i, 1);
  cin >> q;
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int l, r;
      cin >> l >> r;
      l--, r--;
      cout << get(l, r) + 1 << '\n';
    } else {
      int x, y;
      cin >> x >> y;
      x = a[x - 1], y = a[y - 1];
      swap(a[p[x]], a[p[y]]), swap(p[x], p[y]);
      if (x > 0) {
        if (p[x] > p[x - 1])
          modify(x - 1, 0);
        else
          modify(x - 1, 1);
      }
      if (y > 0) {
        if (p[y] > p[y - 1])
          modify(y - 1, 0);
        else
          modify(y - 1, 1);
      }
      if (x < n - 1) {
        if (p[x] < p[x + 1])
          modify(x, 0);
        else
          modify(x, 1);
      }
      if (y < n - 1) {
        if (p[y] < p[y + 1])
          modify(y, 0);
        else
          modify(y, 1);
      }
    }
  }
}

### 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;
int const N = 3e5 + 20;
int n, q, a[N], p[N];
int seg[N << 2];
void modify(int idx, int val, int v = 1, int s = 0, int e = n) {
  if (s + 1 == e) {
    seg[v] = val;
    return;
  }
  int mid = (s + e) >> 1, lc = v << 1, rc = lc | 1;
  if (idx < mid)
    modify(idx, val, lc, s, mid);
  else
    modify(idx, val, rc, mid, e);
  seg[v] = seg[lc] + seg[rc];
}
int get(int l, int r, int v = 1, int s = 0, int e = n) {
  if (s >= r || e <= l) return 0;
  if (s >= l && e <= r) return seg[v];
  int mid = (s + e) >> 1, lc = v << 1, rc = lc | 1;
  return get(l, r, lc, s, mid) + get(l, r, rc, mid, e);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], a[i]--, p[a[i]] = i;
  for (int i = 0; i < n - 1; i++)
    if (p[i] > p[i + 1]) modify(i, 1);
  cin >> q;
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int l, r;
      cin >> l >> r;
      l--, r--;
      cout << get(l, r) + 1 << '\n';
    } else {
      int x, y;
      cin >> x >> y;
      x = a[x - 1], y = a[y - 1];
      swap(a[p[x]], a[p[y]]), swap(p[x], p[y]);
      if (x > 0) {
        if (p[x] > p[x - 1])
          modify(x - 1, 0);
        else
          modify(x - 1, 1);
      }
      if (y > 0) {
        if (p[y] > p[y - 1])
          modify(y - 1, 0);
        else
          modify(y - 1, 1);
      }
      if (x < n - 1) {
        if (p[x] < p[x + 1])
          modify(x, 0);
        else
          modify(x, 1);
      }
      if (y < n - 1) {
        if (p[y] < p[y + 1])
          modify(y, 0);
        else
          modify(y, 1);
      }
    }
  }
}

```