#include <bits/stdc++.h>
using namespace std;
const int INF = (int)1e9;
const long long INF64 = (long long)1e18;
const long double eps = 1e-9;
const long double pi = 3.14159265358979323846;
int nu[301000];
int loc[301000];
int arr[1500010];
int ans(int ss, int ee, int s, int e, int now) {
  if (ss == s && ee == e) return arr[now];
  if (ss == ee) return 1;
  int mid = (s + e) / 2;
  if (ee <= mid) return ans(ss, ee, s, mid, now + now);
  if (ss > mid) return ans(ss, ee, mid + 1, e, now + now + 1);
  int ta = ans(ss, mid, s, mid, now + now);
  int tb = ans(mid + 1, ee, mid + 1, e, now + now + 1);
  int aaans = ta + tb;
  if (loc[mid] < loc[mid + 1]) aaans--;
  return aaans;
}
void build(int ss, int ee, int now) {
  if (ss == ee) {
    arr[now] = 1;
    return;
  }
  int mid = (ss + ee) / 2;
  build(ss, mid, now + now);
  build(mid + 1, ee, now + now + 1);
  arr[now] = arr[now + now] + arr[now + now + 1];
  if (loc[mid] < loc[mid + 1]) arr[now]--;
}
void add(int pos, int ss, int ee, int now) {
  if (ss == ee) return;
  int mid = (ss + ee) / 2;
  if (pos <= mid) add(pos, ss, mid, now + now);
  if (pos > mid) add(pos, mid + 1, ee, now + now + 1);
  arr[now] = arr[now + now] + arr[now + now + 1];
  if (loc[mid] < loc[mid + 1]) arr[now]--;
}
int main() {
  int i;
  int n, qn;
  int q, a, b;
  while (cin >> n) {
    for (i = 0; i < n; i++) {
      cin >> nu[i];
      nu[i]--;
      loc[nu[i]] = i;
    }
    build(0, n - 1, 1);
    cin >> qn;
    while (qn--) {
      cin >> q >> a >> b;
      a--, b--;
      if (q == 2) {
        swap(loc[nu[a]], loc[nu[b]]);
        swap(nu[a], nu[b]);
        add(nu[a], 0, n - 1, 1);
        add(nu[b], 0, n - 1, 1);
      } else {
        printf("%d\n", ans(min(a, b), max(a, b), 0, n - 1, 1));
      }
    }
  }
  return 0;
}

### Prompt

Your task is to create 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 INF = (int)1e9;
const long long INF64 = (long long)1e18;
const long double eps = 1e-9;
const long double pi = 3.14159265358979323846;
int nu[301000];
int loc[301000];
int arr[1500010];
int ans(int ss, int ee, int s, int e, int now) {
  if (ss == s && ee == e) return arr[now];
  if (ss == ee) return 1;
  int mid = (s + e) / 2;
  if (ee <= mid) return ans(ss, ee, s, mid, now + now);
  if (ss > mid) return ans(ss, ee, mid + 1, e, now + now + 1);
  int ta = ans(ss, mid, s, mid, now + now);
  int tb = ans(mid + 1, ee, mid + 1, e, now + now + 1);
  int aaans = ta + tb;
  if (loc[mid] < loc[mid + 1]) aaans--;
  return aaans;
}
void build(int ss, int ee, int now) {
  if (ss == ee) {
    arr[now] = 1;
    return;
  }
  int mid = (ss + ee) / 2;
  build(ss, mid, now + now);
  build(mid + 1, ee, now + now + 1);
  arr[now] = arr[now + now] + arr[now + now + 1];
  if (loc[mid] < loc[mid + 1]) arr[now]--;
}
void add(int pos, int ss, int ee, int now) {
  if (ss == ee) return;
  int mid = (ss + ee) / 2;
  if (pos <= mid) add(pos, ss, mid, now + now);
  if (pos > mid) add(pos, mid + 1, ee, now + now + 1);
  arr[now] = arr[now + now] + arr[now + now + 1];
  if (loc[mid] < loc[mid + 1]) arr[now]--;
}
int main() {
  int i;
  int n, qn;
  int q, a, b;
  while (cin >> n) {
    for (i = 0; i < n; i++) {
      cin >> nu[i];
      nu[i]--;
      loc[nu[i]] = i;
    }
    build(0, n - 1, 1);
    cin >> qn;
    while (qn--) {
      cin >> q >> a >> b;
      a--, b--;
      if (q == 2) {
        swap(loc[nu[a]], loc[nu[b]]);
        swap(nu[a], nu[b]);
        add(nu[a], 0, n - 1, 1);
        add(nu[b], 0, n - 1, 1);
      } else {
        printf("%d\n", ans(min(a, b), max(a, b), 0, n - 1, 1));
      }
    }
  }
  return 0;
}

```