#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> perm, rperm;
vector<int> tree;
int build(int from, int to, int i) {
  if (from == to) {
    tree[i] = 1;
    return 1;
  }
  int mid = (from + to) / 2;
  int left = build(from, mid, i * 2);
  int right = build(mid + 1, to, i * 2 + 1);
  int cur = rperm[mid] < rperm[mid + 1] ? -1 : 0;
  cur += left + right;
  tree[i] = cur;
  return cur;
}
int query(int from, int to, int qfrom, int qto, int i) {
  if (from > qto || to < qfrom) return 0;
  if (from >= qfrom && to <= qto) return tree[i];
  int mid = (from + to) / 2;
  int left = query(from, mid, qfrom, qto, i * 2);
  int right = query(mid + 1, to, qfrom, qto, i * 2 + 1);
  int cur = left + right;
  if (left > 0 && right > 0) {
    cur += rperm[mid] < rperm[mid + 1] ? -1 : 0;
  }
  return cur;
}
int update(int from, int to, int l, int r, int i) {
  if (from > r || to < l) return tree[i];
  if (from > l && to < r) return tree[i];
  if (from == to) return 1;
  int mid = (from + to) / 2;
  int left = update(from, mid, l, r, i * 2);
  int right = update(mid + 1, to, l, r, i * 2 + 1);
  int cur = left + right;
  cur += rperm[mid] < rperm[mid + 1] ? -1 : 0;
  tree[i] = cur;
  return cur;
}
int main() {
  scanf("%d", &n);
  perm.resize(n + 1);
  rperm.resize(n + 1);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &perm[i]);
    rperm[perm[i]] = i;
  }
  int size = (int)(pow(2, ceil(log(n) / log(2)) + 1) + 1);
  tree.resize(size);
  build(1, n, 1);
  int q, type, a, b;
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    scanf("%d %d %d", &type, &a, &b);
    if (type == 1) {
      printf("%d\n", query(1, n, min(a, b), max(a, b), 1));
    } else if (type == 2) {
      swap(rperm[perm[a]], rperm[perm[b]]);
      swap(perm[a], perm[b]);
      update(1, n, min(perm[a], perm[b]), max(perm[a], perm[b]), 1);
    }
  }
  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;
int n;
vector<int> perm, rperm;
vector<int> tree;
int build(int from, int to, int i) {
  if (from == to) {
    tree[i] = 1;
    return 1;
  }
  int mid = (from + to) / 2;
  int left = build(from, mid, i * 2);
  int right = build(mid + 1, to, i * 2 + 1);
  int cur = rperm[mid] < rperm[mid + 1] ? -1 : 0;
  cur += left + right;
  tree[i] = cur;
  return cur;
}
int query(int from, int to, int qfrom, int qto, int i) {
  if (from > qto || to < qfrom) return 0;
  if (from >= qfrom && to <= qto) return tree[i];
  int mid = (from + to) / 2;
  int left = query(from, mid, qfrom, qto, i * 2);
  int right = query(mid + 1, to, qfrom, qto, i * 2 + 1);
  int cur = left + right;
  if (left > 0 && right > 0) {
    cur += rperm[mid] < rperm[mid + 1] ? -1 : 0;
  }
  return cur;
}
int update(int from, int to, int l, int r, int i) {
  if (from > r || to < l) return tree[i];
  if (from > l && to < r) return tree[i];
  if (from == to) return 1;
  int mid = (from + to) / 2;
  int left = update(from, mid, l, r, i * 2);
  int right = update(mid + 1, to, l, r, i * 2 + 1);
  int cur = left + right;
  cur += rperm[mid] < rperm[mid + 1] ? -1 : 0;
  tree[i] = cur;
  return cur;
}
int main() {
  scanf("%d", &n);
  perm.resize(n + 1);
  rperm.resize(n + 1);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &perm[i]);
    rperm[perm[i]] = i;
  }
  int size = (int)(pow(2, ceil(log(n) / log(2)) + 1) + 1);
  tree.resize(size);
  build(1, n, 1);
  int q, type, a, b;
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    scanf("%d %d %d", &type, &a, &b);
    if (type == 1) {
      printf("%d\n", query(1, n, min(a, b), max(a, b), 1));
    } else if (type == 2) {
      swap(rperm[perm[a]], rperm[perm[b]]);
      swap(perm[a], perm[b]);
      update(1, n, min(perm[a], perm[b]), max(perm[a], perm[b]), 1);
    }
  }
  return 0;
}

```