#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 29;
const double EPS = 1e-8;
const double PI = acos(-1);
const int MAX_N = 200000;
struct SegSum {
  int bit[MAX_N + 1];
  int n;
  void init(int _n) {
    n = _n;
    memset(bit, 0, sizeof(bit));
  }
  int sum(int i) {
    int s = 0;
    while (i > 0) {
      s += bit[i];
      i -= i & -i;
    }
    return s;
  }
  int sum(int a, int b) { return sum(b) - sum(a - 1); }
  void add(int i, int x) {
    while (i <= n) {
      bit[i] += x;
      i += i & -i;
    }
  }
} SegSum;
int a[300000];
int pos[300000];
int main() {
  int n;
  while (cin >> n) {
    for (int i = 0; i < (int)n; ++i) {
      scanf("%d", a + i);
      a[i]--;
      pos[a[i]] = i;
    }
    SegSum.init(n);
    for (int i = 0; i < (int)n; ++i) {
      if (a[i] == 0 || pos[a[i] - 1] > i) {
        SegSum.add(a[i] + 1, 1);
      }
    }
    int q;
    cin >> q;
    for (int i = 0; i < (int)q; ++i) {
      int k, x, y;
      scanf("%d%d%d", &k, &x, &y);
      x--;
      y--;
      if (k == 1) {
        int r = SegSum.sum(x + 1, y + 1);
        if (SegSum.sum(x + 1, x + 1) == 0) r++;
        cout << r << endl;
      } else {
        if (a[x] + 1 < n && pos[a[x] + 1] > x && pos[a[x] + 1] < y) {
          SegSum.add(a[x] + 1 + 1, 1);
        }
        if (a[x] - 1 >= 0 && pos[a[x] - 1] > x && pos[a[x] - 1] < y) {
          SegSum.add(a[x] + 1, -1);
        }
        if (a[y] + 1 < n && pos[a[y] + 1] >= x && pos[a[y] + 1] < y) {
          SegSum.add(a[y] + 1 + 1, -1);
        }
        if (a[y] - 1 >= 0 && pos[a[y] - 1] >= x && pos[a[y] - 1] < y) {
          SegSum.add(a[y] + 1, 1);
        }
        swap(a[x], a[y]);
        swap(pos[a[x]], pos[a[y]]);
      }
    }
  }
}

### Prompt

Generate a CPP solution 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;
const int INF = 1 << 29;
const double EPS = 1e-8;
const double PI = acos(-1);
const int MAX_N = 200000;
struct SegSum {
  int bit[MAX_N + 1];
  int n;
  void init(int _n) {
    n = _n;
    memset(bit, 0, sizeof(bit));
  }
  int sum(int i) {
    int s = 0;
    while (i > 0) {
      s += bit[i];
      i -= i & -i;
    }
    return s;
  }
  int sum(int a, int b) { return sum(b) - sum(a - 1); }
  void add(int i, int x) {
    while (i <= n) {
      bit[i] += x;
      i += i & -i;
    }
  }
} SegSum;
int a[300000];
int pos[300000];
int main() {
  int n;
  while (cin >> n) {
    for (int i = 0; i < (int)n; ++i) {
      scanf("%d", a + i);
      a[i]--;
      pos[a[i]] = i;
    }
    SegSum.init(n);
    for (int i = 0; i < (int)n; ++i) {
      if (a[i] == 0 || pos[a[i] - 1] > i) {
        SegSum.add(a[i] + 1, 1);
      }
    }
    int q;
    cin >> q;
    for (int i = 0; i < (int)q; ++i) {
      int k, x, y;
      scanf("%d%d%d", &k, &x, &y);
      x--;
      y--;
      if (k == 1) {
        int r = SegSum.sum(x + 1, y + 1);
        if (SegSum.sum(x + 1, x + 1) == 0) r++;
        cout << r << endl;
      } else {
        if (a[x] + 1 < n && pos[a[x] + 1] > x && pos[a[x] + 1] < y) {
          SegSum.add(a[x] + 1 + 1, 1);
        }
        if (a[x] - 1 >= 0 && pos[a[x] - 1] > x && pos[a[x] - 1] < y) {
          SegSum.add(a[x] + 1, -1);
        }
        if (a[y] + 1 < n && pos[a[y] + 1] >= x && pos[a[y] + 1] < y) {
          SegSum.add(a[y] + 1 + 1, -1);
        }
        if (a[y] - 1 >= 0 && pos[a[y] - 1] >= x && pos[a[y] - 1] < y) {
          SegSum.add(a[y] + 1, 1);
        }
        swap(a[x], a[y]);
        swap(pos[a[x]], pos[a[y]]);
      }
    }
  }
}

```