#include <bits/stdc++.h>
using namespace std;
long long rdtsc() {
  long long tmp;
  asm("rdtsc" : "=A"(tmp));
  return tmp;
}
inline int myrand() { return abs((rand() << 15) ^ rand()); }
inline int rnd(int x) { return myrand() % x; }
const int INF = (int)1e9 + 1;
const long double EPS = 1e-9;
const int maxn = (int)3e5 + 10;
int a[maxn], inv[maxn];
struct FenvTree {
  int a[maxn];
  int n;
  FenvTree(int _n) : n(_n) {
    for (int i = 0; i < n; ++i) a[i] = 0;
  }
  void add(int pos, int toadd) {
    for (int x = pos; x < n; x |= (x + 1)) a[x] += toadd;
  }
  int get(int pos) const {
    int res = 0;
    for (int x = pos; x >= 0; x = (x & (x + 1)) - 1) res += a[x];
    return res;
  }
};
int n;
void make(int pos, int nval, FenvTree &tree) {
  if (pos < n - 1)
    tree.add(pos, (int)(nval > inv[pos + 1]) - (int)(inv[pos] > inv[pos + 1]));
  if (pos > 0)
    tree.add(pos - 1,
             (int)(inv[pos - 1] > nval) - (int)(inv[pos - 1] > inv[pos]));
  inv[pos] = nval;
}
bool solve() {
  if (scanf("%d", &n) < 1) return 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    --a[i];
    inv[a[i]] = i;
  }
  int q;
  assert(scanf("%d", &q) == 1);
  FenvTree tree(n - 1);
  for (int i = 0; i < n; ++i) tree.add(i, (int)(inv[i] > inv[i + 1]));
  for (int iter = 0; iter < q; ++iter) {
    int p, x, y;
    scanf("%d%d%d", &p, &x, &y);
    --x, --y;
    if (p == 1) {
      int res = 1 + tree.get(y - 1) - tree.get(x - 1);
      printf("%d\n", res);
      continue;
    }
    if (p == 2) {
      int p1 = a[x], p2 = a[y];
      make(p1, y, tree);
      make(p2, x, tree);
      swap(a[x], a[y]);
      continue;
    }
    assert(0);
  }
  return 1;
}
int main() {
  srand(rdtsc());
  while (1) {
    if (!solve()) break;
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
using namespace std;
long long rdtsc() {
  long long tmp;
  asm("rdtsc" : "=A"(tmp));
  return tmp;
}
inline int myrand() { return abs((rand() << 15) ^ rand()); }
inline int rnd(int x) { return myrand() % x; }
const int INF = (int)1e9 + 1;
const long double EPS = 1e-9;
const int maxn = (int)3e5 + 10;
int a[maxn], inv[maxn];
struct FenvTree {
  int a[maxn];
  int n;
  FenvTree(int _n) : n(_n) {
    for (int i = 0; i < n; ++i) a[i] = 0;
  }
  void add(int pos, int toadd) {
    for (int x = pos; x < n; x |= (x + 1)) a[x] += toadd;
  }
  int get(int pos) const {
    int res = 0;
    for (int x = pos; x >= 0; x = (x & (x + 1)) - 1) res += a[x];
    return res;
  }
};
int n;
void make(int pos, int nval, FenvTree &tree) {
  if (pos < n - 1)
    tree.add(pos, (int)(nval > inv[pos + 1]) - (int)(inv[pos] > inv[pos + 1]));
  if (pos > 0)
    tree.add(pos - 1,
             (int)(inv[pos - 1] > nval) - (int)(inv[pos - 1] > inv[pos]));
  inv[pos] = nval;
}
bool solve() {
  if (scanf("%d", &n) < 1) return 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    --a[i];
    inv[a[i]] = i;
  }
  int q;
  assert(scanf("%d", &q) == 1);
  FenvTree tree(n - 1);
  for (int i = 0; i < n; ++i) tree.add(i, (int)(inv[i] > inv[i + 1]));
  for (int iter = 0; iter < q; ++iter) {
    int p, x, y;
    scanf("%d%d%d", &p, &x, &y);
    --x, --y;
    if (p == 1) {
      int res = 1 + tree.get(y - 1) - tree.get(x - 1);
      printf("%d\n", res);
      continue;
    }
    if (p == 2) {
      int p1 = a[x], p2 = a[y];
      make(p1, y, tree);
      make(p2, x, tree);
      swap(a[x], a[y]);
      continue;
    }
    assert(0);
  }
  return 1;
}
int main() {
  srand(rdtsc());
  while (1) {
    if (!solve()) break;
  }
  return 0;
}

```