#include <bits/stdc++.h>
using namespace std;
const int inf = 1.01e9;
const double eps = 1e-9;
struct fenwik {
  vector<long long> t;
  fenwik() {}
  fenwik(int n) { t = vector<long long>(n); }
  long long get(int x) {
    x += 1;
    long long res = 0;
    for (; x > 0; x -= x & -x) res += t[x];
    return res;
  }
  void upd(int x, int dval) {
    x += 1;
    for (; x < ((int)(t).size()); x += x & -x) t[x] += dval;
  }
};
const int N = 3e5 + 10;
int a[N];
int a1[N];
fenwik f;
int b[N];
int n;
void upd(int x) {
  if (x < 0) return;
  if (x >= n - 1) return;
  int v = a1[x] > a1[x + 1];
  if (b[x] != v) {
    f.upd(x, v - b[x]);
    b[x] = v;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; (i) < (n); ++i) scanf("%d", &a[i]);
  for (int i = 0; (i) < (n); ++i) a[i]--;
  f = fenwik(n + 10);
  for (int i = 0; (i) < (n); ++i) a1[a[i]] = i;
  for (int i = 0; (i) < (n - 1); ++i) upd(i);
  int q;
  scanf("%d", &q);
  for (int _ = 0; (_) < (q); ++_) {
    int tp;
    scanf("%d", &tp);
    if (tp == 1) {
      int x, y;
      scanf("%d%d", &x, &y);
      --x, --y;
      int res = f.get(y - 1) - f.get(x - 1) + 1;
      printf("%d\n", res);
    } else {
      int x, y;
      scanf("%d%d", &x, &y);
      --x, --y;
      swap(a[x], a[y]);
      swap(a1[a[x]], a1[a[y]]);
      upd(a[x] - 1);
      upd(a[x]);
      upd(a[y] - 1);
      upd(a[y]);
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 inf = 1.01e9;
const double eps = 1e-9;
struct fenwik {
  vector<long long> t;
  fenwik() {}
  fenwik(int n) { t = vector<long long>(n); }
  long long get(int x) {
    x += 1;
    long long res = 0;
    for (; x > 0; x -= x & -x) res += t[x];
    return res;
  }
  void upd(int x, int dval) {
    x += 1;
    for (; x < ((int)(t).size()); x += x & -x) t[x] += dval;
  }
};
const int N = 3e5 + 10;
int a[N];
int a1[N];
fenwik f;
int b[N];
int n;
void upd(int x) {
  if (x < 0) return;
  if (x >= n - 1) return;
  int v = a1[x] > a1[x + 1];
  if (b[x] != v) {
    f.upd(x, v - b[x]);
    b[x] = v;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; (i) < (n); ++i) scanf("%d", &a[i]);
  for (int i = 0; (i) < (n); ++i) a[i]--;
  f = fenwik(n + 10);
  for (int i = 0; (i) < (n); ++i) a1[a[i]] = i;
  for (int i = 0; (i) < (n - 1); ++i) upd(i);
  int q;
  scanf("%d", &q);
  for (int _ = 0; (_) < (q); ++_) {
    int tp;
    scanf("%d", &tp);
    if (tp == 1) {
      int x, y;
      scanf("%d%d", &x, &y);
      --x, --y;
      int res = f.get(y - 1) - f.get(x - 1) + 1;
      printf("%d\n", res);
    } else {
      int x, y;
      scanf("%d%d", &x, &y);
      --x, --y;
      swap(a[x], a[y]);
      swap(a1[a[x]], a1[a[y]]);
      upd(a[x] - 1);
      upd(a[x]);
      upd(a[y] - 1);
      upd(a[y]);
    }
  }
  return 0;
}

```