#include <bits/stdc++.h>
using namespace std;
int n, a[300005], b[300005], an = 0, t;
int tree[300005], vis[5][300005];
void setIO(string a) {
  string in = a + ".in", out = a + ".out";
  freopen(in.c_str(), "r", stdin);
  freopen(out.c_str(), "w", stdout);
}
void add(int x, int delta) {
  for (int p = x; p <= n; p += p & (-p)) tree[p] += delta;
}
int ask(int x) {
  int an = 0;
  for (int p = x; p; p -= p & (-p)) an += tree[p];
  return an;
}
void mod(int x, int y, int z) {
  if (vis[z][x] == t) return;
  add(x, y);
  vis[z][x] = t;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) {
    scanf("%d", &a[i]);
    b[a[i]] = i;
  }
  for (int i = (1); i <= (n); i++)
    if (b[i] < b[i - 1]) add(i, 1);
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int opt, l, r;
    scanf("%d%d%d", &opt, &l, &r);
    if (opt == 1) {
      if (l > r) swap(l, r);
      printf("%d\n", ask(r) - ask(l) + 1);
    } else {
      ++t;
      int x = a[l], y = a[r];
      if (b[x] < b[x - 1]) mod(x, -1, 0);
      if (b[x + 1] < b[x]) mod(x + 1, -1, 0);
      if (b[y] < b[y - 1]) mod(y, -1, 0);
      if (b[y + 1] < b[y]) mod(y + 1, -1, 0);
      swap(a[l], a[r]);
      swap(b[x], b[y]);
      if (b[x] < b[x - 1]) mod(x, 1, 1);
      if (b[x + 1] < b[x]) mod(x + 1, 1, 1);
      if (b[y] < b[y - 1]) mod(y, 1, 1);
      if (b[y + 1] < b[y]) mod(y + 1, 1, 1);
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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 n, a[300005], b[300005], an = 0, t;
int tree[300005], vis[5][300005];
void setIO(string a) {
  string in = a + ".in", out = a + ".out";
  freopen(in.c_str(), "r", stdin);
  freopen(out.c_str(), "w", stdout);
}
void add(int x, int delta) {
  for (int p = x; p <= n; p += p & (-p)) tree[p] += delta;
}
int ask(int x) {
  int an = 0;
  for (int p = x; p; p -= p & (-p)) an += tree[p];
  return an;
}
void mod(int x, int y, int z) {
  if (vis[z][x] == t) return;
  add(x, y);
  vis[z][x] = t;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) {
    scanf("%d", &a[i]);
    b[a[i]] = i;
  }
  for (int i = (1); i <= (n); i++)
    if (b[i] < b[i - 1]) add(i, 1);
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int opt, l, r;
    scanf("%d%d%d", &opt, &l, &r);
    if (opt == 1) {
      if (l > r) swap(l, r);
      printf("%d\n", ask(r) - ask(l) + 1);
    } else {
      ++t;
      int x = a[l], y = a[r];
      if (b[x] < b[x - 1]) mod(x, -1, 0);
      if (b[x + 1] < b[x]) mod(x + 1, -1, 0);
      if (b[y] < b[y - 1]) mod(y, -1, 0);
      if (b[y + 1] < b[y]) mod(y + 1, -1, 0);
      swap(a[l], a[r]);
      swap(b[x], b[y]);
      if (b[x] < b[x - 1]) mod(x, 1, 1);
      if (b[x + 1] < b[x]) mod(x + 1, 1, 1);
      if (b[y] < b[y - 1]) mod(y, 1, 1);
      if (b[y + 1] < b[y]) mod(y + 1, 1, 1);
    }
  }
  return 0;
}

```