#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 2;
int seg[4 * maxn], a[maxn], p[maxn];
int build(int id, int L, int R) {
  if (R == L + 1) return 0;
  int mid = (L + R) / 2;
  int x = build(2 * id, L, mid), y = build(2 * id + 1, mid, R);
  seg[id] = x + y;
  if (a[mid] < a[mid - 1]) seg[id]++;
  return seg[id];
}
int get(int id, int L, int R, int l, int r) {
  if (R <= l || r <= L) return 0;
  if (L >= l && R <= r) {
    return seg[id];
  }
  int mid = (L + R) / 2, ans;
  int x = get(2 * id, L, mid, l, r), y = get(2 * id + 1, mid, R, l, r);
  ans = x + y;
  if (mid - 1 >= l && mid < r && a[mid] < a[mid - 1]) ans++;
  return ans;
}
int add(int id, int L, int R, int idx, int val) {
  if (idx < L || idx >= R) return seg[id];
  if (R == L + 1) {
    a[idx] = val;
    return 0;
  }
  int mid = (L + R) / 2;
  int x = add(2 * id, L, mid, idx, val), y = add(2 * id + 1, mid, R, idx, val);
  seg[id] = x + y;
  if (a[mid] < a[mid - 1]) seg[id]++;
  return seg[id];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
    a[p[i]] = i;
  }
  build(1, 0, n);
  int q;
  cin >> q;
  while (q--) {
    int type, x, y;
    cin >> type >> x >> y;
    x--;
    y--;
    if (type == 1) {
      cout << get(1, 0, n, x, y + 1) + 1 << '\n';
    } else {
      swap(p[x], p[y]);
      int X = a[p[x]], Y = a[p[y]];
      add(1, 0, n, p[x], Y);
      add(1, 0, n, p[y], X);
    }
  }
}

### Prompt

Generate 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 + 2;
int seg[4 * maxn], a[maxn], p[maxn];
int build(int id, int L, int R) {
  if (R == L + 1) return 0;
  int mid = (L + R) / 2;
  int x = build(2 * id, L, mid), y = build(2 * id + 1, mid, R);
  seg[id] = x + y;
  if (a[mid] < a[mid - 1]) seg[id]++;
  return seg[id];
}
int get(int id, int L, int R, int l, int r) {
  if (R <= l || r <= L) return 0;
  if (L >= l && R <= r) {
    return seg[id];
  }
  int mid = (L + R) / 2, ans;
  int x = get(2 * id, L, mid, l, r), y = get(2 * id + 1, mid, R, l, r);
  ans = x + y;
  if (mid - 1 >= l && mid < r && a[mid] < a[mid - 1]) ans++;
  return ans;
}
int add(int id, int L, int R, int idx, int val) {
  if (idx < L || idx >= R) return seg[id];
  if (R == L + 1) {
    a[idx] = val;
    return 0;
  }
  int mid = (L + R) / 2;
  int x = add(2 * id, L, mid, idx, val), y = add(2 * id + 1, mid, R, idx, val);
  seg[id] = x + y;
  if (a[mid] < a[mid - 1]) seg[id]++;
  return seg[id];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
    a[p[i]] = i;
  }
  build(1, 0, n);
  int q;
  cin >> q;
  while (q--) {
    int type, x, y;
    cin >> type >> x >> y;
    x--;
    y--;
    if (type == 1) {
      cout << get(1, 0, n, x, y + 1) + 1 << '\n';
    } else {
      swap(p[x], p[y]);
      int X = a[p[x]], Y = a[p[y]];
      add(1, 0, n, p[x], Y);
      add(1, 0, n, p[y], X);
    }
  }
}

```