#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
  if (b > a) return gcd(b, a);
  return (b == 0) ? a : gcd(b, a % b);
}
int bitcount(long long a) {
  int i;
  for (i = 0; a > 0; i++) a &= (a - 1);
  return i;
}
int a[300005], b[4 * 300005], d[3 * 100005];
void tree(int n, int s, int e) {
  if (s == e) {
    if (a[s] < a[s - 1])
      b[n] = 1;
    else
      b[n] = 0;
    return;
  }
  int m = (s + e) / 2;
  tree(2 * n, s, m);
  tree(2 * n + 1, m + 1, e);
  b[n] = b[2 * n] + b[2 * n + 1];
  return;
}
int query(int n, int s, int e, int p1, int p2) {
  if (s > p2 || e < p1) return 0;
  if (s >= p1 && e <= p2) return b[n];
  int m = (s + e) / 2;
  return query(2 * n, s, m, p1, p2) + query(2 * n + 1, m + 1, e, p1, p2);
}
void update(int n, int s, int e, int p) {
  if (s == p && e == p) {
    if (a[s] < a[s - 1])
      b[n] = 1;
    else
      b[n] = 0;
    return;
  }
  int m = (s + e) / 2;
  if (p <= m)
    update(2 * n, s, m, p);
  else
    update(2 * n + 1, m + 1, e, p);
  b[n] = b[2 * n] + b[2 * n + 1];
  return;
}
int main() {
  int n, q, i, j, k;
  cin >> n;
  a[0] = 0;
  for (i = 1; i <= n; i++) {
    cin >> d[i];
    a[d[i]] = i;
  }
  tree(1, 1, n);
  cin >> q;
  while (q--) {
    cin >> i >> j >> k;
    if (i == 1) {
      cout << query(1, 1, n, j + 1, k) + 1 << endl;
    } else {
      swap(d[j], d[k]);
      a[d[j]] = j;
      a[d[k]] = k;
      update(1, 1, n, d[j]);
      update(1, 1, n, d[k]);
      if (d[k] != n) update(1, 1, n, d[k] + 1);
      if (d[j] != n) update(1, 1, n, d[j] + 1);
    }
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
int gcd(int a, int b) {
  if (b > a) return gcd(b, a);
  return (b == 0) ? a : gcd(b, a % b);
}
int bitcount(long long a) {
  int i;
  for (i = 0; a > 0; i++) a &= (a - 1);
  return i;
}
int a[300005], b[4 * 300005], d[3 * 100005];
void tree(int n, int s, int e) {
  if (s == e) {
    if (a[s] < a[s - 1])
      b[n] = 1;
    else
      b[n] = 0;
    return;
  }
  int m = (s + e) / 2;
  tree(2 * n, s, m);
  tree(2 * n + 1, m + 1, e);
  b[n] = b[2 * n] + b[2 * n + 1];
  return;
}
int query(int n, int s, int e, int p1, int p2) {
  if (s > p2 || e < p1) return 0;
  if (s >= p1 && e <= p2) return b[n];
  int m = (s + e) / 2;
  return query(2 * n, s, m, p1, p2) + query(2 * n + 1, m + 1, e, p1, p2);
}
void update(int n, int s, int e, int p) {
  if (s == p && e == p) {
    if (a[s] < a[s - 1])
      b[n] = 1;
    else
      b[n] = 0;
    return;
  }
  int m = (s + e) / 2;
  if (p <= m)
    update(2 * n, s, m, p);
  else
    update(2 * n + 1, m + 1, e, p);
  b[n] = b[2 * n] + b[2 * n + 1];
  return;
}
int main() {
  int n, q, i, j, k;
  cin >> n;
  a[0] = 0;
  for (i = 1; i <= n; i++) {
    cin >> d[i];
    a[d[i]] = i;
  }
  tree(1, 1, n);
  cin >> q;
  while (q--) {
    cin >> i >> j >> k;
    if (i == 1) {
      cout << query(1, 1, n, j + 1, k) + 1 << endl;
    } else {
      swap(d[j], d[k]);
      a[d[j]] = j;
      a[d[k]] = k;
      update(1, 1, n, d[j]);
      update(1, 1, n, d[k]);
      if (d[k] != n) update(1, 1, n, d[k] + 1);
      if (d[j] != n) update(1, 1, n, d[j] + 1);
    }
  }
  return 0;
}

```