#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
int D[N], A[N], IND[N], n, q, tp;
void add(int ind) {
  for (ind++; ind < N; ind += (ind & (-ind))) D[ind]++;
}
void del(int ind) {
  for (ind++; ind < N; ind += (ind & (-ind))) D[ind]--;
}
int get(int ind) {
  int sm = 0;
  for (ind++; ind; ind -= (ind & (-ind))) sm += D[ind];
  return (sm);
}
int get(int l, int r) { return (get(r) - get(l - 1)); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  for (int i = 1; i <= n; i++) IND[A[i]] = i;
  for (int i = 2; i <= n; i++)
    if (IND[i] > IND[i - 1]) add(i);
  scanf("%d", &q);
  for (int Q = 0; Q < q; Q++) {
    int x, y;
    scanf("%d %d %d", &tp, &x, &y);
    if (tp == 1) {
      printf("%d\n", y - x + 1 - get(x + 1, y));
      continue;
    }
    int a = A[x];
    vector<int> st1;
    st1.push_back(-1);
    if (a > 1 && IND[a] > IND[a - 1]) st1.push_back(a);
    if (a < n && IND[a] < IND[a + 1]) st1.push_back(a + 1);
    a = A[y];
    if (a > 1 && IND[a] > IND[a - 1]) st1.push_back(a);
    if (a < n && IND[a] < IND[a + 1]) st1.push_back(a + 1);
    sort(st1.begin(), st1.end());
    for (int i = 1; i < (int)st1.size(); i++)
      if (st1[i] ^ st1[i - 1]) del(st1[i]);
    swap(A[x], A[y]);
    IND[A[x]] = x;
    IND[A[y]] = y;
    vector<int> st2;
    st2.push_back(-1);
    a = A[x];
    if (a > 1 && IND[a] > IND[a - 1]) st2.push_back(a);
    if (a < n && IND[a] < IND[a + 1]) st2.push_back(a + 1);
    a = A[y];
    if (a > 1 && IND[a] > IND[a - 1]) st2.push_back(a);
    if (a < n && IND[a] < IND[a + 1]) st2.push_back(a + 1);
    sort(st2.begin(), st2.end());
    for (int i = 1; i < (int)st2.size(); i++)
      if (st2[i] ^ st2[i - 1]) add(st2[i]);
  }
}

### 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;
const int N = 3e5 + 5;
int D[N], A[N], IND[N], n, q, tp;
void add(int ind) {
  for (ind++; ind < N; ind += (ind & (-ind))) D[ind]++;
}
void del(int ind) {
  for (ind++; ind < N; ind += (ind & (-ind))) D[ind]--;
}
int get(int ind) {
  int sm = 0;
  for (ind++; ind; ind -= (ind & (-ind))) sm += D[ind];
  return (sm);
}
int get(int l, int r) { return (get(r) - get(l - 1)); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  for (int i = 1; i <= n; i++) IND[A[i]] = i;
  for (int i = 2; i <= n; i++)
    if (IND[i] > IND[i - 1]) add(i);
  scanf("%d", &q);
  for (int Q = 0; Q < q; Q++) {
    int x, y;
    scanf("%d %d %d", &tp, &x, &y);
    if (tp == 1) {
      printf("%d\n", y - x + 1 - get(x + 1, y));
      continue;
    }
    int a = A[x];
    vector<int> st1;
    st1.push_back(-1);
    if (a > 1 && IND[a] > IND[a - 1]) st1.push_back(a);
    if (a < n && IND[a] < IND[a + 1]) st1.push_back(a + 1);
    a = A[y];
    if (a > 1 && IND[a] > IND[a - 1]) st1.push_back(a);
    if (a < n && IND[a] < IND[a + 1]) st1.push_back(a + 1);
    sort(st1.begin(), st1.end());
    for (int i = 1; i < (int)st1.size(); i++)
      if (st1[i] ^ st1[i - 1]) del(st1[i]);
    swap(A[x], A[y]);
    IND[A[x]] = x;
    IND[A[y]] = y;
    vector<int> st2;
    st2.push_back(-1);
    a = A[x];
    if (a > 1 && IND[a] > IND[a - 1]) st2.push_back(a);
    if (a < n && IND[a] < IND[a + 1]) st2.push_back(a + 1);
    a = A[y];
    if (a > 1 && IND[a] > IND[a - 1]) st2.push_back(a);
    if (a < n && IND[a] < IND[a + 1]) st2.push_back(a + 1);
    sort(st2.begin(), st2.end());
    for (int i = 1; i < (int)st2.size(); i++)
      if (st2[i] ^ st2[i - 1]) add(st2[i]);
  }
}

```