#include <bits/stdc++.h>
using namespace std;
const long long A = 100000000000000LL, N = 300228;
long long t[N * 4], i, j, n, m, a[N], p[N], o, b, l, r;
void modi(long long v, long long l, long long r, long long p, long long g) {
  if (p < l || p > r) return;
  if (l == r) {
    t[v] = g;
    return;
  }
  long long mid = (l + r) / 2;
  if (p <= mid)
    modi(v * 2, l, mid, p, g);
  else
    modi(v * 2 + 1, mid + 1, r, p, g);
  t[v] = t[v * 2] + t[v * 2 + 1];
}
long long get(long long v, long long l, long long r, long long _l,
              long long _r) {
  if (_l > _r) return 0;
  if (l == _l && r == _r) return t[v];
  long long mid = (l + r) / 2;
  return get(v * 2, l, mid, _l, min(_r, mid)) +
         get(v * 2 + 1, mid + 1, r, max(mid + 1, _l), _r);
}
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), p[a[i]] = i;
  for (i = 1; i < n; i++)
    if (p[i] > p[i + 1]) modi(1, 1, n, i + 1, 1);
  cin >> m;
  while (m--) {
    scanf("%d%d%d", &b, &l, &r);
    if (b == 1)
      printf("%d\n", get(1, 1, n, l + 1, r) + 1);
    else {
      swap(p[a[l]], p[a[r]]), swap(a[l], a[r]);
      if (p[a[l]] > p[a[l] + 1])
        modi(1, 1, n, a[l] + 1, 1);
      else
        modi(1, 1, n, a[l] + 1, 0);
      if (p[a[r]] > p[a[r] + 1])
        modi(1, 1, n, a[r] + 1, 1);
      else
        modi(1, 1, n, a[r] + 1, 0);
      if (p[a[l] - 1] > p[a[l]])
        modi(1, 1, n, a[l], 1);
      else
        modi(1, 1, n, a[l], 0);
      if (p[a[r] - 1] > p[a[r]])
        modi(1, 1, n, a[r], 1);
      else
        modi(1, 1, n, a[r], 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;
const long long A = 100000000000000LL, N = 300228;
long long t[N * 4], i, j, n, m, a[N], p[N], o, b, l, r;
void modi(long long v, long long l, long long r, long long p, long long g) {
  if (p < l || p > r) return;
  if (l == r) {
    t[v] = g;
    return;
  }
  long long mid = (l + r) / 2;
  if (p <= mid)
    modi(v * 2, l, mid, p, g);
  else
    modi(v * 2 + 1, mid + 1, r, p, g);
  t[v] = t[v * 2] + t[v * 2 + 1];
}
long long get(long long v, long long l, long long r, long long _l,
              long long _r) {
  if (_l > _r) return 0;
  if (l == _l && r == _r) return t[v];
  long long mid = (l + r) / 2;
  return get(v * 2, l, mid, _l, min(_r, mid)) +
         get(v * 2 + 1, mid + 1, r, max(mid + 1, _l), _r);
}
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), p[a[i]] = i;
  for (i = 1; i < n; i++)
    if (p[i] > p[i + 1]) modi(1, 1, n, i + 1, 1);
  cin >> m;
  while (m--) {
    scanf("%d%d%d", &b, &l, &r);
    if (b == 1)
      printf("%d\n", get(1, 1, n, l + 1, r) + 1);
    else {
      swap(p[a[l]], p[a[r]]), swap(a[l], a[r]);
      if (p[a[l]] > p[a[l] + 1])
        modi(1, 1, n, a[l] + 1, 1);
      else
        modi(1, 1, n, a[l] + 1, 0);
      if (p[a[r]] > p[a[r] + 1])
        modi(1, 1, n, a[r] + 1, 1);
      else
        modi(1, 1, n, a[r] + 1, 0);
      if (p[a[l] - 1] > p[a[l]])
        modi(1, 1, n, a[l], 1);
      else
        modi(1, 1, n, a[l], 0);
      if (p[a[r] - 1] > p[a[r]])
        modi(1, 1, n, a[r], 1);
      else
        modi(1, 1, n, a[r], 0);
    }
  }
}

```