#include <bits/stdc++.h>
using namespace std;
const int Maxn = 300005;
const int Maxm = 2097152;
int n;
int a[Maxn], pos[Maxn];
int res[Maxm];
int q;
void Create(int v, int l, int r) {
  if (l == r)
    res[v] = 1;
  else {
    int m = l + r >> 1;
    Create(2 * v, l, m);
    Create(2 * v + 1, m + 1, r);
    res[v] = res[2 * v] - (pos[m] < pos[m + 1]) + res[2 * v + 1];
  }
}
int Get(int v, int l, int r, int a, int b) {
  if (l == a && r == b)
    return res[v];
  else {
    int m = l + r >> 1;
    if (b <= m) return Get(2 * v, l, m, a, b);
    if (m + 1 <= a) return Get(2 * v + 1, m + 1, r, a, b);
    return Get(2 * v, l, m, a, m) - (pos[m] < pos[m + 1]) +
           Get(2 * v + 1, m + 1, r, m + 1, b);
  }
}
void Update(int v, int l, int r, int x) {
  if (l == r)
    res[v] = 1;
  else {
    int m = l + r >> 1;
    if (x <= m)
      Update(2 * v, l, m, x);
    else
      Update(2 * v + 1, m + 1, r, x);
    res[v] = res[2 * v] - (pos[m] < pos[m + 1]) + res[2 * v + 1];
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  Create(1, 1, n);
  scanf("%d", &q);
  while (q--) {
    int p, x, y;
    scanf("%d %d %d", &p, &x, &y);
    if (p == 1)
      printf("%d\n", Get(1, 1, n, x, y));
    else {
      pos[a[x - 1]] = y - 1;
      pos[a[y - 1]] = x - 1;
      swap(a[x - 1], a[y - 1]);
      Update(1, 1, n, a[x - 1]);
      Update(1, 1, n, a[y - 1]);
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 = 300005;
const int Maxm = 2097152;
int n;
int a[Maxn], pos[Maxn];
int res[Maxm];
int q;
void Create(int v, int l, int r) {
  if (l == r)
    res[v] = 1;
  else {
    int m = l + r >> 1;
    Create(2 * v, l, m);
    Create(2 * v + 1, m + 1, r);
    res[v] = res[2 * v] - (pos[m] < pos[m + 1]) + res[2 * v + 1];
  }
}
int Get(int v, int l, int r, int a, int b) {
  if (l == a && r == b)
    return res[v];
  else {
    int m = l + r >> 1;
    if (b <= m) return Get(2 * v, l, m, a, b);
    if (m + 1 <= a) return Get(2 * v + 1, m + 1, r, a, b);
    return Get(2 * v, l, m, a, m) - (pos[m] < pos[m + 1]) +
           Get(2 * v + 1, m + 1, r, m + 1, b);
  }
}
void Update(int v, int l, int r, int x) {
  if (l == r)
    res[v] = 1;
  else {
    int m = l + r >> 1;
    if (x <= m)
      Update(2 * v, l, m, x);
    else
      Update(2 * v + 1, m + 1, r, x);
    res[v] = res[2 * v] - (pos[m] < pos[m + 1]) + res[2 * v + 1];
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  Create(1, 1, n);
  scanf("%d", &q);
  while (q--) {
    int p, x, y;
    scanf("%d %d %d", &p, &x, &y);
    if (p == 1)
      printf("%d\n", Get(1, 1, n, x, y));
    else {
      pos[a[x - 1]] = y - 1;
      pos[a[y - 1]] = x - 1;
      swap(a[x - 1], a[y - 1]);
      Update(1, 1, n, a[x - 1]);
      Update(1, 1, n, a[y - 1]);
    }
  }
  return 0;
}

```