#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:640000000")
using namespace std;
const long long INF = (1e15) + 4;
const int N = 3e5 + 5;
int a[N];
int b[N];
int p[N];
int n;
int len;
int t[4 * N];
void upd(int idx, int l, int r, int pos) {
  if (l == r) {
    t[idx] = 1;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid) {
    upd(idx * 2 + 1, l, mid, pos);
  } else {
    upd(idx * 2 + 2, mid + 1, r, pos);
  }
  t[idx] = t[idx * 2 + 1] + t[idx * 2 + 2];
  if (a[mid] < a[mid + 1]) t[idx]--;
}
int get(int idx, int l, int r, int L, int R) {
  if (l == L && r == R) {
    return t[idx];
  }
  int mid = (l + r) >> 1;
  int ans = 0;
  int cnt = 0;
  if (L <= mid) {
    ans += get(idx * 2 + 1, l, mid, L, min(mid, R));
    cnt++;
  }
  if (R >= mid + 1) {
    ans += get(idx * 2 + 2, mid + 1, r, max(mid + 1, L), R);
    cnt++;
  }
  if (cnt == 2) ans -= (a[mid] < a[mid + 1]);
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int first;
    scanf("%d", &first);
    first--;
    a[first] = i;
    p[i] = first;
  }
  for (int i = 0; i < n; i++) {
    upd(0, 0, n - 1, i);
  }
  int m;
  scanf("%d", &m);
  int last = n + 1;
  for (int i = 0; i < m; i++) {
    int type, l, r;
    scanf("%d %d %d", &type, &l, &r);
    l--;
    r--;
    if (type == 1) {
      printf("%d\n", get(0, 0, n - 1, l, r));
    } else {
      swap(p[l], p[r]);
      swap(a[p[l]], a[p[r]]);
      upd(0, 0, n - 1, p[l]);
      upd(0, 0, n - 1, p[r]);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma comment(linker, "/STACK:640000000")
using namespace std;
const long long INF = (1e15) + 4;
const int N = 3e5 + 5;
int a[N];
int b[N];
int p[N];
int n;
int len;
int t[4 * N];
void upd(int idx, int l, int r, int pos) {
  if (l == r) {
    t[idx] = 1;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid) {
    upd(idx * 2 + 1, l, mid, pos);
  } else {
    upd(idx * 2 + 2, mid + 1, r, pos);
  }
  t[idx] = t[idx * 2 + 1] + t[idx * 2 + 2];
  if (a[mid] < a[mid + 1]) t[idx]--;
}
int get(int idx, int l, int r, int L, int R) {
  if (l == L && r == R) {
    return t[idx];
  }
  int mid = (l + r) >> 1;
  int ans = 0;
  int cnt = 0;
  if (L <= mid) {
    ans += get(idx * 2 + 1, l, mid, L, min(mid, R));
    cnt++;
  }
  if (R >= mid + 1) {
    ans += get(idx * 2 + 2, mid + 1, r, max(mid + 1, L), R);
    cnt++;
  }
  if (cnt == 2) ans -= (a[mid] < a[mid + 1]);
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int first;
    scanf("%d", &first);
    first--;
    a[first] = i;
    p[i] = first;
  }
  for (int i = 0; i < n; i++) {
    upd(0, 0, n - 1, i);
  }
  int m;
  scanf("%d", &m);
  int last = n + 1;
  for (int i = 0; i < m; i++) {
    int type, l, r;
    scanf("%d %d %d", &type, &l, &r);
    l--;
    r--;
    if (type == 1) {
      printf("%d\n", get(0, 0, n - 1, l, r));
    } else {
      swap(p[l], p[r]);
      swap(a[p[l]], a[p[r]]);
      upd(0, 0, n - 1, p[l]);
      upd(0, 0, n - 1, p[r]);
    }
  }
  return 0;
}

```