#include <bits/stdc++.h>
using namespace std;
const int MAXN = 300010;
struct node {
  int l, r, sum;
} t[MAXN << 2];
void build(int l, int r, int k) {
  t[k].l = l;
  t[k].r = r;
  t[k].sum = 0;
  if (l == r) return;
  int mid = (t[k].l + t[k].r) >> 1;
  build(l, mid, k << 1);
  build(mid + 1, r, k << 1 | 1);
}
void update(int k) { t[k].sum = t[k << 1].sum + t[k << 1 | 1].sum; }
void insert(int pos, int val, int k) {
  if (t[k].l == t[k].r) {
    t[k].sum = val;
    return;
  }
  int mid = (t[k].l + t[k].r) >> 1;
  if (pos <= mid)
    insert(pos, val, k << 1);
  else
    insert(pos, val, k << 1 | 1);
  update(k);
}
int query(int l, int r, int k) {
  if (l <= t[k].l && t[k].r <= r) return t[k].sum;
  int mid = (t[k].l + t[k].r) >> 1;
  if (r <= mid)
    return query(l, r, k << 1);
  else if (l >= mid + 1)
    return query(l, r, k << 1 | 1);
  else
    return query(l, mid, k << 1) + query(mid + 1, r, k << 1 | 1);
}
int a[MAXN], p[MAXN];
int n;
void gao(int v, int pos) {
  if (v > 1) {
    if (p[v - 1] < pos)
      insert(v - 1, 1, 1);
    else
      insert(v - 1, 0, 1);
  }
  if (v < n) {
    if (pos < p[v + 1])
      insert(v, 1, 1);
    else
      insert(v, 0, 1);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++) p[a[i]] = i;
  build(1, n - 1, 1);
  for (int i = 1; i < n; i++) {
    if (p[i] < p[i + 1]) {
      insert(i, 1, 1);
    }
  }
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1)
      printf("%d\n", y - x + 1 - query(x, y - 1, 1));
    else {
      int s = a[x], t = a[y];
      swap(a[x], a[y]);
      swap(p[s], p[t]);
      gao(a[x], x);
      gao(a[y], y);
    }
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 = 300010;
struct node {
  int l, r, sum;
} t[MAXN << 2];
void build(int l, int r, int k) {
  t[k].l = l;
  t[k].r = r;
  t[k].sum = 0;
  if (l == r) return;
  int mid = (t[k].l + t[k].r) >> 1;
  build(l, mid, k << 1);
  build(mid + 1, r, k << 1 | 1);
}
void update(int k) { t[k].sum = t[k << 1].sum + t[k << 1 | 1].sum; }
void insert(int pos, int val, int k) {
  if (t[k].l == t[k].r) {
    t[k].sum = val;
    return;
  }
  int mid = (t[k].l + t[k].r) >> 1;
  if (pos <= mid)
    insert(pos, val, k << 1);
  else
    insert(pos, val, k << 1 | 1);
  update(k);
}
int query(int l, int r, int k) {
  if (l <= t[k].l && t[k].r <= r) return t[k].sum;
  int mid = (t[k].l + t[k].r) >> 1;
  if (r <= mid)
    return query(l, r, k << 1);
  else if (l >= mid + 1)
    return query(l, r, k << 1 | 1);
  else
    return query(l, mid, k << 1) + query(mid + 1, r, k << 1 | 1);
}
int a[MAXN], p[MAXN];
int n;
void gao(int v, int pos) {
  if (v > 1) {
    if (p[v - 1] < pos)
      insert(v - 1, 1, 1);
    else
      insert(v - 1, 0, 1);
  }
  if (v < n) {
    if (pos < p[v + 1])
      insert(v, 1, 1);
    else
      insert(v, 0, 1);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++) p[a[i]] = i;
  build(1, n - 1, 1);
  for (int i = 1; i < n; i++) {
    if (p[i] < p[i + 1]) {
      insert(i, 1, 1);
    }
  }
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1)
      printf("%d\n", y - x + 1 - query(x, y - 1, 1));
    else {
      int s = a[x], t = a[y];
      swap(a[x], a[y]);
      swap(p[s], p[t]);
      gao(a[x], x);
      gao(a[y], y);
    }
  }
  return 0;
}

```