#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3 * 100100;
int T[4 * MAXN];
int pow2(int n) {
  int q = 1;
  while (q < n) q *= 2;
  return q;
}
int pos[MAXN];
int perm[MAXN];
int n, P;
int build(int i) {
  if (i >= P) return T[i];
  return T[i] = build(2 * i) + build(2 * i + 1);
}
void init(int b, int e) {
  b = max(1, b);
  e = min(e, n);
  for (int i = b; i <= e; i++)
    if (pos[i + 1] < pos[i])
      T[P + i - 1] = 1;
    else
      T[P + i - 1] = 0;
}
int get(int v, int b, int e, int B, int E) {
  if (e == b) return 0;
  if (e <= B || b >= E) return 0;
  if (b <= B && e >= E) return T[v];
  return get(2 * v, b, e, B, (B + E) / 2) +
         get(2 * v + 1, b, e, (B + E) / 2, E);
}
int check(int a, int b) {
  a--;
  b--;
  return get(1, a, b, 0, P);
}
void update(int v) {
  if (v == 0) return;
  T[v] = T[2 * v] + T[2 * v + 1];
  update(v / 2);
}
void doUpdate(int b, int e) {
  b = max(b, 1);
  e = min(e, n);
  b--;
  e--;
  for (int i = b; i <= e; i++) update((i + P) / 2);
}
void swp(int a, int b) {
  int i = perm[a], j = perm[b];
  swap(perm[a], perm[b]);
  swap(pos[i], pos[j]);
  init(i - 1, i + 1);
  init(j - 1, j + 1);
  doUpdate(i - 1, i + 1);
  doUpdate(j - 1, j + 1);
}
int main() {
  scanf("%d", &n);
  P = pow2(n);
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    pos[a] = i;
    perm[i] = a;
  }
  pos[n + 1] = n + 1;
  perm[n + 1] = n + 1;
  init(1, n);
  build(1);
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int ot, a, b;
    scanf("%d %d %d", &ot, &a, &b);
    if (ot == 1)
      printf("%d\n", 1 + check(a, b));
    else
      swp(a, b);
  }
  return 0;
}

### Prompt

Generate a Cpp solution to 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;
const int MAXN = 3 * 100100;
int T[4 * MAXN];
int pow2(int n) {
  int q = 1;
  while (q < n) q *= 2;
  return q;
}
int pos[MAXN];
int perm[MAXN];
int n, P;
int build(int i) {
  if (i >= P) return T[i];
  return T[i] = build(2 * i) + build(2 * i + 1);
}
void init(int b, int e) {
  b = max(1, b);
  e = min(e, n);
  for (int i = b; i <= e; i++)
    if (pos[i + 1] < pos[i])
      T[P + i - 1] = 1;
    else
      T[P + i - 1] = 0;
}
int get(int v, int b, int e, int B, int E) {
  if (e == b) return 0;
  if (e <= B || b >= E) return 0;
  if (b <= B && e >= E) return T[v];
  return get(2 * v, b, e, B, (B + E) / 2) +
         get(2 * v + 1, b, e, (B + E) / 2, E);
}
int check(int a, int b) {
  a--;
  b--;
  return get(1, a, b, 0, P);
}
void update(int v) {
  if (v == 0) return;
  T[v] = T[2 * v] + T[2 * v + 1];
  update(v / 2);
}
void doUpdate(int b, int e) {
  b = max(b, 1);
  e = min(e, n);
  b--;
  e--;
  for (int i = b; i <= e; i++) update((i + P) / 2);
}
void swp(int a, int b) {
  int i = perm[a], j = perm[b];
  swap(perm[a], perm[b]);
  swap(pos[i], pos[j]);
  init(i - 1, i + 1);
  init(j - 1, j + 1);
  doUpdate(i - 1, i + 1);
  doUpdate(j - 1, j + 1);
}
int main() {
  scanf("%d", &n);
  P = pow2(n);
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    pos[a] = i;
    perm[i] = a;
  }
  pos[n + 1] = n + 1;
  perm[n + 1] = n + 1;
  init(1, n);
  build(1);
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int ot, a, b;
    scanf("%d %d %d", &ot, &a, &b);
    if (ot == 1)
      printf("%d\n", 1 + check(a, b));
    else
      swp(a, b);
  }
  return 0;
}

```