#include <bits/stdc++.h>
using namespace std;
struct FenwickTree {
  int c[1100000], n;
  void init(int _n) {
    n = _n;
    memset(c, 0, sizeof(int) * (n + 1));
  }
  void add(int x, int val) {
    while (x <= n) {
      c[x] += val;
      x += (x & (-x));
    }
  }
  int sum(int x) {
    int res = 0;
    while (x) {
      res += c[x];
      x -= (x & (-x));
    }
    return res;
  }
} ft;
int n, pos[1100000], a[1100000];
int sgn(int x) { return x > 0 ? 1 : -1; }
void cao(int x, int y) {
  int v = a[x], w;
  if (v != 1) {
    w = pos[v - 1];
    int d = sgn(y - w) - sgn(x - w);
    ft.add(v, (-d) >> 1);
  }
  if (v != n) {
    w = pos[v + 1];
    int d = sgn(y - w) - sgn(x - w);
    ft.add(v + 1, d >> 1);
  }
}
void cao1(int x, int y) {
  int v = a[x], w;
  if (v != 1) {
    w = pos[v - 1];
    int d = sgn(y - w) - sgn(x - w);
    ft.add(v, (-d) >> 1);
  }
}
void cao2(int x, int y) {
  int v = a[x], w;
  if (v != n) {
    w = pos[v + 1];
    int d = sgn(y - w) - sgn(x - w);
    ft.add(v + 1, d >> 1);
  }
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), pos[a[i]] = i;
    ft.init(n);
    ft.add(1, 1);
    for (int i = 2; i <= n; i++) {
      if (pos[i] < pos[i - 1]) ft.add(i, 1);
    }
    int q;
    scanf("%d", &q);
    while (q--) {
      int op, l, r;
      scanf("%d%d%d", &op, &l, &r);
      if (op == 1) {
        printf("%d\n", ft.sum(r) - ft.sum(l) + 1);
      } else {
        if (abs(a[l] - a[r]) > 1) {
          cao(l, r);
          cao(r, l);
        } else {
          if (a[l] > a[r]) swap(l, r);
          if (r > l)
            ft.add(a[r], 1);
          else
            ft.add(a[r], -1);
          cao1(l, r);
          cao2(r, l);
        }
        swap(a[l], a[r]);
        swap(pos[a[l]], pos[a[r]]);
      }
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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;
struct FenwickTree {
  int c[1100000], n;
  void init(int _n) {
    n = _n;
    memset(c, 0, sizeof(int) * (n + 1));
  }
  void add(int x, int val) {
    while (x <= n) {
      c[x] += val;
      x += (x & (-x));
    }
  }
  int sum(int x) {
    int res = 0;
    while (x) {
      res += c[x];
      x -= (x & (-x));
    }
    return res;
  }
} ft;
int n, pos[1100000], a[1100000];
int sgn(int x) { return x > 0 ? 1 : -1; }
void cao(int x, int y) {
  int v = a[x], w;
  if (v != 1) {
    w = pos[v - 1];
    int d = sgn(y - w) - sgn(x - w);
    ft.add(v, (-d) >> 1);
  }
  if (v != n) {
    w = pos[v + 1];
    int d = sgn(y - w) - sgn(x - w);
    ft.add(v + 1, d >> 1);
  }
}
void cao1(int x, int y) {
  int v = a[x], w;
  if (v != 1) {
    w = pos[v - 1];
    int d = sgn(y - w) - sgn(x - w);
    ft.add(v, (-d) >> 1);
  }
}
void cao2(int x, int y) {
  int v = a[x], w;
  if (v != n) {
    w = pos[v + 1];
    int d = sgn(y - w) - sgn(x - w);
    ft.add(v + 1, d >> 1);
  }
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), pos[a[i]] = i;
    ft.init(n);
    ft.add(1, 1);
    for (int i = 2; i <= n; i++) {
      if (pos[i] < pos[i - 1]) ft.add(i, 1);
    }
    int q;
    scanf("%d", &q);
    while (q--) {
      int op, l, r;
      scanf("%d%d%d", &op, &l, &r);
      if (op == 1) {
        printf("%d\n", ft.sum(r) - ft.sum(l) + 1);
      } else {
        if (abs(a[l] - a[r]) > 1) {
          cao(l, r);
          cao(r, l);
        } else {
          if (a[l] > a[r]) swap(l, r);
          if (r > l)
            ft.add(a[r], 1);
          else
            ft.add(a[r], -1);
          cao1(l, r);
          cao2(r, l);
        }
        swap(a[l], a[r]);
        swap(pos[a[l]], pos[a[r]]);
      }
    }
  }
  return 0;
}

```