#include <bits/stdc++.h>
using namespace std;
int n;
int a[300010];
int id[300010];
int tree[4 * 300010];
int last[4 * 300010];
int first[4 * 300010];
void swap(int &x, int &y) {
  int tmp = x;
  x = y;
  y = tmp;
}
void init(int ind, int l, int r) {
  if (l == r) {
    tree[ind] = 0;
    first[ind] = id[l];
    last[ind] = id[l];
    return;
  }
  int h = (l + r) / 2;
  init(2 * ind, l, h);
  init(2 * ind + 1, h + 1, r);
  tree[ind] = tree[2 * ind] + tree[2 * ind + 1];
  first[ind] = first[2 * ind];
  last[ind] = last[2 * ind + 1];
  if (first[2 * ind + 1] < last[2 * ind]) tree[ind]++;
}
void update(int ind, int l, int r, int t, int v) {
  if (l > t || r < t) return;
  if (l == r && l == t) {
    first[ind] = v;
    last[ind] = v;
    return;
  }
  int h = (l + r) / 2;
  update(2 * ind, l, h, t, v);
  update(2 * ind + 1, h + 1, r, t, v);
  tree[ind] = tree[2 * ind] + tree[2 * ind + 1];
  first[ind] = first[2 * ind];
  last[ind] = last[2 * ind + 1];
  if (first[2 * ind + 1] < last[2 * ind]) tree[ind]++;
}
int get(int ind, int l, int r, int x, int y) {
  if (l > y || r < x) return 0;
  if (l >= x && r <= y) return tree[ind];
  int tmp = 0;
  int h = (l + r) / 2;
  tmp += get(2 * ind, l, h, x, y);
  tmp += get(2 * ind + 1, h + 1, r, x, y);
  if (h >= x && (h + 1) <= y && first[2 * ind + 1] < last[2 * ind]) tmp++;
  return tmp;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  for (int i = 0; i < n; i++) id[a[i]] = i + 1;
  init(1, 1, n);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int t;
    scanf("%d", &t);
    if (t == 2) {
      int u, v;
      scanf("%d%d", &u, &v);
      u--;
      v--;
      int tmp = id[a[u]];
      update(1, 1, n, a[u], id[a[v]]);
      id[a[u]] = id[a[v]];
      update(1, 1, n, a[v], tmp);
      id[a[v]] = tmp;
      swap(a[u], a[v]);
    } else {
      int x, y;
      scanf("%d%d", &x, &y);
      int tot = 0;
      tot = get(1, 1, n, x, y) + 1;
      printf("%d\n", tot);
    }
  }
  return 0;
}

### 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 n;
int a[300010];
int id[300010];
int tree[4 * 300010];
int last[4 * 300010];
int first[4 * 300010];
void swap(int &x, int &y) {
  int tmp = x;
  x = y;
  y = tmp;
}
void init(int ind, int l, int r) {
  if (l == r) {
    tree[ind] = 0;
    first[ind] = id[l];
    last[ind] = id[l];
    return;
  }
  int h = (l + r) / 2;
  init(2 * ind, l, h);
  init(2 * ind + 1, h + 1, r);
  tree[ind] = tree[2 * ind] + tree[2 * ind + 1];
  first[ind] = first[2 * ind];
  last[ind] = last[2 * ind + 1];
  if (first[2 * ind + 1] < last[2 * ind]) tree[ind]++;
}
void update(int ind, int l, int r, int t, int v) {
  if (l > t || r < t) return;
  if (l == r && l == t) {
    first[ind] = v;
    last[ind] = v;
    return;
  }
  int h = (l + r) / 2;
  update(2 * ind, l, h, t, v);
  update(2 * ind + 1, h + 1, r, t, v);
  tree[ind] = tree[2 * ind] + tree[2 * ind + 1];
  first[ind] = first[2 * ind];
  last[ind] = last[2 * ind + 1];
  if (first[2 * ind + 1] < last[2 * ind]) tree[ind]++;
}
int get(int ind, int l, int r, int x, int y) {
  if (l > y || r < x) return 0;
  if (l >= x && r <= y) return tree[ind];
  int tmp = 0;
  int h = (l + r) / 2;
  tmp += get(2 * ind, l, h, x, y);
  tmp += get(2 * ind + 1, h + 1, r, x, y);
  if (h >= x && (h + 1) <= y && first[2 * ind + 1] < last[2 * ind]) tmp++;
  return tmp;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  for (int i = 0; i < n; i++) id[a[i]] = i + 1;
  init(1, 1, n);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int t;
    scanf("%d", &t);
    if (t == 2) {
      int u, v;
      scanf("%d%d", &u, &v);
      u--;
      v--;
      int tmp = id[a[u]];
      update(1, 1, n, a[u], id[a[v]]);
      id[a[u]] = id[a[v]];
      update(1, 1, n, a[v], tmp);
      id[a[v]] = tmp;
      swap(a[u], a[v]);
    } else {
      int x, y;
      scanf("%d%d", &x, &y);
      int tot = 0;
      tot = get(1, 1, n, x, y) + 1;
      printf("%d\n", tot);
    }
  }
  return 0;
}

```