#include <bits/stdc++.h>
const double eps = 1e-9;
const double pi = acos(-1);
const int oo = 1000000000;
const int mod = 1000000007;
const double E = 2.7182818284590452353602874713527;
using namespace std;
int tree[300001], a[300001], b[300001], yes[300001];
int n;
void ins(int i, int x) {
  if (x == 1 && yes[i] || x == -1 && !yes[i]) return;
  if (x == 1)
    yes[i] = 1;
  else
    yes[i] = 0;
  for (; i <= n; i += ((i) & (-(i)))) tree[i] += x;
}
int sum(int i) {
  int s = 0;
  for (; i; i -= ((i) & (-(i)))) s += tree[i];
  return s;
}
int main() {
  int q;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[a[i]] = i;
  }
  for (int i = 2; i <= n; i++)
    if (b[i] < b[i - 1]) ins(i, 1);
  cin >> q;
  while (q--) {
    int t, x, y;
    scanf("%d%d%d", &t, &x, &y);
    if (t == 1)
      printf("%d\n", sum(y) - sum(x) + 1);
    else {
      if (a[x] > 1 && b[a[x]] < b[a[x] - 1]) ins(a[x], -1);
      if (a[x] < n && b[a[x] + 1] < b[a[x]]) ins(a[x] + 1, -1);
      if (a[y] > 1 && b[a[y]] < b[a[y] - 1]) ins(a[y], -1);
      if (a[y] < n && b[a[y] + 1] < b[a[y]]) ins(a[y] + 1, -1);
      a[x] ^= a[y];
      a[y] ^= a[x];
      a[x] ^= a[y];
      b[a[x]] = x;
      b[a[y]] = y;
      if (a[x] > 1 && b[a[x]] < b[a[x] - 1]) ins(a[x], 1);
      if (a[x] < n && b[a[x] + 1] < b[a[x]]) ins(a[x] + 1, 1);
      if (a[y] > 1 && b[a[y]] < b[a[y] - 1]) ins(a[y], 1);
      if (a[y] < n && b[a[y] + 1] < b[a[y]]) ins(a[y] + 1, 1);
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
const double eps = 1e-9;
const double pi = acos(-1);
const int oo = 1000000000;
const int mod = 1000000007;
const double E = 2.7182818284590452353602874713527;
using namespace std;
int tree[300001], a[300001], b[300001], yes[300001];
int n;
void ins(int i, int x) {
  if (x == 1 && yes[i] || x == -1 && !yes[i]) return;
  if (x == 1)
    yes[i] = 1;
  else
    yes[i] = 0;
  for (; i <= n; i += ((i) & (-(i)))) tree[i] += x;
}
int sum(int i) {
  int s = 0;
  for (; i; i -= ((i) & (-(i)))) s += tree[i];
  return s;
}
int main() {
  int q;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[a[i]] = i;
  }
  for (int i = 2; i <= n; i++)
    if (b[i] < b[i - 1]) ins(i, 1);
  cin >> q;
  while (q--) {
    int t, x, y;
    scanf("%d%d%d", &t, &x, &y);
    if (t == 1)
      printf("%d\n", sum(y) - sum(x) + 1);
    else {
      if (a[x] > 1 && b[a[x]] < b[a[x] - 1]) ins(a[x], -1);
      if (a[x] < n && b[a[x] + 1] < b[a[x]]) ins(a[x] + 1, -1);
      if (a[y] > 1 && b[a[y]] < b[a[y] - 1]) ins(a[y], -1);
      if (a[y] < n && b[a[y] + 1] < b[a[y]]) ins(a[y] + 1, -1);
      a[x] ^= a[y];
      a[y] ^= a[x];
      a[x] ^= a[y];
      b[a[x]] = x;
      b[a[y]] = y;
      if (a[x] > 1 && b[a[x]] < b[a[x] - 1]) ins(a[x], 1);
      if (a[x] < n && b[a[x] + 1] < b[a[x]]) ins(a[x] + 1, 1);
      if (a[y] > 1 && b[a[y]] < b[a[y] - 1]) ins(a[y], 1);
      if (a[y] < n && b[a[y] + 1] < b[a[y]]) ins(a[y] + 1, 1);
    }
  }
  return 0;
}

```