#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 50;
const int INF = 0x3f3f3f3f;
int a[maxn], id[maxn];
struct node {
  int l, r, sum;
} tree[maxn << 2];
void push_up(int i) { tree[i].sum = tree[i << 1].sum + tree[i << 1 | 1].sum; }
void build(int i, int l, int r) {
  tree[i].l = l;
  tree[i].r = r;
  tree[i].sum = 0;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build(i << 1, l, mid);
  build(i << 1 | 1, mid + 1, r);
  push_up(i);
}
void updata(int i, int x, int k) {
  if (tree[i].l == tree[i].r && tree[i].l == x) {
    tree[i].sum = k;
    return;
  }
  int mid = (tree[i].l + tree[i].r) >> 1;
  if (x <= mid)
    updata(i << 1, x, k);
  else
    updata(i << 1 | 1, x, k);
  push_up(i);
}
int query(int i, int x, int y) {
  if (tree[i].l >= x && tree[i].r <= y) return tree[i].sum;
  int mid = (tree[i].l + tree[i].r) >> 1;
  if (y <= mid)
    return query(i << 1, x, y);
  else if (x > mid)
    return query(i << 1 | 1, x, y);
  else
    return query(i << 1, x, mid) + query(i << 1 | 1, mid + 1, y);
}
int main() {
  int n;
  scanf("%d", &n);
  memset(id, 0, sizeof(id));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    id[a[i]] = i;
  }
  build(1, 0, maxn - 30);
  for (int i = 1; i <= n; i++)
    if (id[a[i] + 1] < id[a[i]]) updata(1, a[i], 1);
  int m;
  scanf("%d", &m);
  while (m--) {
    int op, l, r;
    scanf("%d%d%d", &op, &l, &r);
    if (op == 1) {
      int ans = query(1, l, r);
      if (query(1, r, r) == 0) ans++;
      printf("%d\n", ans);
    } else {
      swap(id[a[l]], id[a[r]]);
      swap(a[l], a[r]);
      for (int i = -1; i <= 1; i++) {
        int k = a[l] + i;
        if (id[k + 1] < id[k])
          updata(1, k, 1);
        else
          updata(1, k, 0);
        k = a[r] + i;
        if (id[k + 1] < id[k])
          updata(1, k, 1);
        else
          updata(1, k, 0);
      }
    }
  }
}

### Prompt

Please formulate 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 = 3e5 + 50;
const int INF = 0x3f3f3f3f;
int a[maxn], id[maxn];
struct node {
  int l, r, sum;
} tree[maxn << 2];
void push_up(int i) { tree[i].sum = tree[i << 1].sum + tree[i << 1 | 1].sum; }
void build(int i, int l, int r) {
  tree[i].l = l;
  tree[i].r = r;
  tree[i].sum = 0;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build(i << 1, l, mid);
  build(i << 1 | 1, mid + 1, r);
  push_up(i);
}
void updata(int i, int x, int k) {
  if (tree[i].l == tree[i].r && tree[i].l == x) {
    tree[i].sum = k;
    return;
  }
  int mid = (tree[i].l + tree[i].r) >> 1;
  if (x <= mid)
    updata(i << 1, x, k);
  else
    updata(i << 1 | 1, x, k);
  push_up(i);
}
int query(int i, int x, int y) {
  if (tree[i].l >= x && tree[i].r <= y) return tree[i].sum;
  int mid = (tree[i].l + tree[i].r) >> 1;
  if (y <= mid)
    return query(i << 1, x, y);
  else if (x > mid)
    return query(i << 1 | 1, x, y);
  else
    return query(i << 1, x, mid) + query(i << 1 | 1, mid + 1, y);
}
int main() {
  int n;
  scanf("%d", &n);
  memset(id, 0, sizeof(id));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    id[a[i]] = i;
  }
  build(1, 0, maxn - 30);
  for (int i = 1; i <= n; i++)
    if (id[a[i] + 1] < id[a[i]]) updata(1, a[i], 1);
  int m;
  scanf("%d", &m);
  while (m--) {
    int op, l, r;
    scanf("%d%d%d", &op, &l, &r);
    if (op == 1) {
      int ans = query(1, l, r);
      if (query(1, r, r) == 0) ans++;
      printf("%d\n", ans);
    } else {
      swap(id[a[l]], id[a[r]]);
      swap(a[l], a[r]);
      for (int i = -1; i <= 1; i++) {
        int k = a[l] + i;
        if (id[k + 1] < id[k])
          updata(1, k, 1);
        else
          updata(1, k, 0);
        k = a[r] + i;
        if (id[k + 1] < id[k])
          updata(1, k, 1);
        else
          updata(1, k, 0);
      }
    }
  }
}

```