#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-11;
const int inf = 0x3f3f3f3f;
template <class T>
void show(T a, int n) {
  for (int i = 0; i < n; ++i) cout << a[i] << ' ';
  cout << endl;
}
template <class T>
void show(T a, int r, int l) {
  for (int i = 0; i < r; ++i) show(a[i], l);
  cout << endl;
}
const int N = 3e5 + 111;
bool f[N];
int a[N];
int re[N];
int C[N];
inline int lowbit(int x) { return x & -x; }
void add(int x, int d) {
  while (x < N) {
    C[x] += d;
    x += lowbit(x);
  }
}
int sum(int x) {
  int res = 0;
  while (x > 0) {
    res += C[x];
    x -= lowbit(x);
  }
  return res;
}
int main() {
  int n, i, j, q, b, x, y, c, d;
  cin >> n;
  for (i = 1; i <= n; i++) {
    scanf("%d", &b);
    a[b] = i;
    re[i] = b;
  }
  for (i = 1; i <= n; i++) {
    if (a[i] < a[i - 1]) {
      f[i] = 1;
      add(i, 1);
    }
  }
  cin >> q;
  while (q--) {
    scanf("%d%d%d", &b, &x, &y);
    if (b == 1) {
      printf("%d\n", sum(y) - sum(x) + 1);
    } else {
      c = re[x];
      d = re[y];
      a[c] = y;
      a[d] = x;
      re[y] = c;
      re[x] = d;
      if (c > 1) {
        if (a[c - 1] < a[c] && f[c]) {
          f[c] = 0;
          add(c, -1);
        } else if (a[c - 1] > a[c] && !f[c]) {
          f[c] = 1;
          add(c, 1);
        }
      }
      if (c < n) {
        if (a[c] < a[c + 1] && f[c + 1]) {
          f[c + 1] = 0;
          add(c + 1, -1);
        } else if (a[c] > a[c + 1] && !f[c + 1]) {
          f[c + 1] = 1;
          add(c + 1, 1);
        }
      }
      if (d > 1) {
        if (a[d - 1] < a[d] && f[d]) {
          f[d] = 0;
          add(d, -1);
        } else if (a[d - 1] > a[d] && !f[d]) {
          f[d] = 1;
          add(d, 1);
        }
      }
      if (d < n) {
        if (a[d] < a[d + 1] && f[d + 1]) {
          f[d + 1] = 0;
          add(d + 1, -1);
        } else if (a[d] > a[d + 1] && !f[d + 1]) {
          f[d + 1] = 1;
          add(d + 1, 1);
        }
      }
    }
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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>
#pragma warning(disable : 4996)
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-11;
const int inf = 0x3f3f3f3f;
template <class T>
void show(T a, int n) {
  for (int i = 0; i < n; ++i) cout << a[i] << ' ';
  cout << endl;
}
template <class T>
void show(T a, int r, int l) {
  for (int i = 0; i < r; ++i) show(a[i], l);
  cout << endl;
}
const int N = 3e5 + 111;
bool f[N];
int a[N];
int re[N];
int C[N];
inline int lowbit(int x) { return x & -x; }
void add(int x, int d) {
  while (x < N) {
    C[x] += d;
    x += lowbit(x);
  }
}
int sum(int x) {
  int res = 0;
  while (x > 0) {
    res += C[x];
    x -= lowbit(x);
  }
  return res;
}
int main() {
  int n, i, j, q, b, x, y, c, d;
  cin >> n;
  for (i = 1; i <= n; i++) {
    scanf("%d", &b);
    a[b] = i;
    re[i] = b;
  }
  for (i = 1; i <= n; i++) {
    if (a[i] < a[i - 1]) {
      f[i] = 1;
      add(i, 1);
    }
  }
  cin >> q;
  while (q--) {
    scanf("%d%d%d", &b, &x, &y);
    if (b == 1) {
      printf("%d\n", sum(y) - sum(x) + 1);
    } else {
      c = re[x];
      d = re[y];
      a[c] = y;
      a[d] = x;
      re[y] = c;
      re[x] = d;
      if (c > 1) {
        if (a[c - 1] < a[c] && f[c]) {
          f[c] = 0;
          add(c, -1);
        } else if (a[c - 1] > a[c] && !f[c]) {
          f[c] = 1;
          add(c, 1);
        }
      }
      if (c < n) {
        if (a[c] < a[c + 1] && f[c + 1]) {
          f[c + 1] = 0;
          add(c + 1, -1);
        } else if (a[c] > a[c + 1] && !f[c + 1]) {
          f[c + 1] = 1;
          add(c + 1, 1);
        }
      }
      if (d > 1) {
        if (a[d - 1] < a[d] && f[d]) {
          f[d] = 0;
          add(d, -1);
        } else if (a[d - 1] > a[d] && !f[d]) {
          f[d] = 1;
          add(d, 1);
        }
      }
      if (d < n) {
        if (a[d] < a[d + 1] && f[d + 1]) {
          f[d + 1] = 0;
          add(d + 1, -1);
        } else if (a[d] > a[d + 1] && !f[d + 1]) {
          f[d + 1] = 1;
          add(d + 1, 1);
        }
      }
    }
  }
  return 0;
}

```