#include <bits/stdc++.h>
using namespace std;
int ans[300008], pos[300008], a[300008];
struct Tree {
  int left, right, value;
} tree[1200100];
void build(int l, int r, int id) {
  tree[id].left = l;
  tree[id].right = r;
  if (l == r)
    tree[id].value = ans[l];
  else {
    int mid = (l + r) / 2;
    build(l, mid, 2 * id);
    build(mid + 1, r, 2 * id + 1);
    tree[id].value = tree[2 * id].value + tree[2 * id + 1].value;
  }
}
void update(int x, int id, int val) {
  if (tree[id].left == tree[id].right)
    tree[id].value = val;
  else {
    int mid = (tree[id].left + tree[id].right) / 2;
    if (x > mid)
      update(x, 2 * id + 1, val);
    else
      update(x, 2 * id, val);
    tree[id].value = tree[2 * id].value + tree[2 * id + 1].value;
  }
}
int query(int l, int r, int id) {
  if (tree[id].left == l && tree[id].right == r)
    return tree[id].value;
  else {
    int mid = (tree[id].left + tree[id].right) / 2;
    if (l > mid)
      return query(l, r, 2 * id + 1);
    else if (r <= mid)
      return query(l, r, 2 * id);
    else
      return query(l, mid, 2 * id) + query(mid + 1, r, 2 * id + 1);
  }
}
int main() {
  int n, q, flag, x, y, i, j, temp1, temp2;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    a[x] = i;
    pos[i] = x;
  }
  for (i = 1; i <= n;) {
    ans[i] = 1;
    for (j = i + 1; j <= n && a[j] > a[j - 1]; j++) ans[j] = 0;
    i = j;
  }
  build(1, n, 1);
  scanf("%d", &q);
  for (i = 0; i < q; i++) {
    scanf("%d%d%d", &flag, &x, &y);
    if (flag == 1) {
      if (ans[x])
        printf("%d\n", query(x, y, 1));
      else
        printf("%d\n", query(x, y, 1) + 1);
    } else {
      temp1 = pos[x];
      temp2 = pos[y];
      if (temp1 > 1 && (y > a[temp1 - 1] || y == a[temp1 - 1]) &&
          x < a[temp1 - 1]) {
        update(temp1, 1, 0);
        ans[temp1] = 0;
      } else if (temp1 > 1 && (y < a[temp1 - 1] || y == a[temp1 - 1]) &&
                 x > a[temp1 - 1]) {
        update(temp1, 1, 1);
        ans[temp1] = 1;
      }
      if (temp1 < n && (y > a[temp1 + 1] || y == a[temp1 + 1]) &&
          x < a[temp1 + 1]) {
        update(temp1 + 1, 1, 1);
        ans[temp1 + 1] = 1;
      } else if (temp1 < n && (y < a[temp1 + 1] || y == a[temp1 + 1]) &&
                 x > a[temp1 + 1]) {
        update(temp1 + 1, 1, 0);
        ans[temp1 + 1] = 0;
      }
      if (temp2 > 1 && temp2 != temp1 + 1 &&
          (x > a[temp2 - 1] || x == a[temp2 - 1]) && y < a[temp2 - 1]) {
        update(temp2, 1, 0);
        ans[temp2] = 0;
      } else if (temp2 > 1 && temp2 != temp1 + 1 &&
                 (x < a[temp2 - 1] || x == a[temp2 - 1]) && y > a[temp2 - 1]) {
        update(temp2, 1, 1);
        ans[temp2] = 1;
      }
      if (temp2 < n && temp2 != temp1 - 1 &&
          (x > a[temp2 + 1] || x == a[temp2 + 1]) && y < a[temp2 + 1]) {
        update(temp2 + 1, 1, 1);
        ans[temp2 + 1] = 1;
      } else if (temp2 < n && temp2 != temp1 - 1 &&
                 (x < a[temp2 + 1] || x == a[temp2 + 1]) && y > a[temp2 + 1]) {
        update(temp2 + 1, 1, 0);
        ans[temp2 + 1] = 0;
      }
      a[temp1] = y;
      a[temp2] = x;
      pos[x] = pos[y];
      pos[y] = temp1;
    }
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 ans[300008], pos[300008], a[300008];
struct Tree {
  int left, right, value;
} tree[1200100];
void build(int l, int r, int id) {
  tree[id].left = l;
  tree[id].right = r;
  if (l == r)
    tree[id].value = ans[l];
  else {
    int mid = (l + r) / 2;
    build(l, mid, 2 * id);
    build(mid + 1, r, 2 * id + 1);
    tree[id].value = tree[2 * id].value + tree[2 * id + 1].value;
  }
}
void update(int x, int id, int val) {
  if (tree[id].left == tree[id].right)
    tree[id].value = val;
  else {
    int mid = (tree[id].left + tree[id].right) / 2;
    if (x > mid)
      update(x, 2 * id + 1, val);
    else
      update(x, 2 * id, val);
    tree[id].value = tree[2 * id].value + tree[2 * id + 1].value;
  }
}
int query(int l, int r, int id) {
  if (tree[id].left == l && tree[id].right == r)
    return tree[id].value;
  else {
    int mid = (tree[id].left + tree[id].right) / 2;
    if (l > mid)
      return query(l, r, 2 * id + 1);
    else if (r <= mid)
      return query(l, r, 2 * id);
    else
      return query(l, mid, 2 * id) + query(mid + 1, r, 2 * id + 1);
  }
}
int main() {
  int n, q, flag, x, y, i, j, temp1, temp2;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    a[x] = i;
    pos[i] = x;
  }
  for (i = 1; i <= n;) {
    ans[i] = 1;
    for (j = i + 1; j <= n && a[j] > a[j - 1]; j++) ans[j] = 0;
    i = j;
  }
  build(1, n, 1);
  scanf("%d", &q);
  for (i = 0; i < q; i++) {
    scanf("%d%d%d", &flag, &x, &y);
    if (flag == 1) {
      if (ans[x])
        printf("%d\n", query(x, y, 1));
      else
        printf("%d\n", query(x, y, 1) + 1);
    } else {
      temp1 = pos[x];
      temp2 = pos[y];
      if (temp1 > 1 && (y > a[temp1 - 1] || y == a[temp1 - 1]) &&
          x < a[temp1 - 1]) {
        update(temp1, 1, 0);
        ans[temp1] = 0;
      } else if (temp1 > 1 && (y < a[temp1 - 1] || y == a[temp1 - 1]) &&
                 x > a[temp1 - 1]) {
        update(temp1, 1, 1);
        ans[temp1] = 1;
      }
      if (temp1 < n && (y > a[temp1 + 1] || y == a[temp1 + 1]) &&
          x < a[temp1 + 1]) {
        update(temp1 + 1, 1, 1);
        ans[temp1 + 1] = 1;
      } else if (temp1 < n && (y < a[temp1 + 1] || y == a[temp1 + 1]) &&
                 x > a[temp1 + 1]) {
        update(temp1 + 1, 1, 0);
        ans[temp1 + 1] = 0;
      }
      if (temp2 > 1 && temp2 != temp1 + 1 &&
          (x > a[temp2 - 1] || x == a[temp2 - 1]) && y < a[temp2 - 1]) {
        update(temp2, 1, 0);
        ans[temp2] = 0;
      } else if (temp2 > 1 && temp2 != temp1 + 1 &&
                 (x < a[temp2 - 1] || x == a[temp2 - 1]) && y > a[temp2 - 1]) {
        update(temp2, 1, 1);
        ans[temp2] = 1;
      }
      if (temp2 < n && temp2 != temp1 - 1 &&
          (x > a[temp2 + 1] || x == a[temp2 + 1]) && y < a[temp2 + 1]) {
        update(temp2 + 1, 1, 1);
        ans[temp2 + 1] = 1;
      } else if (temp2 < n && temp2 != temp1 - 1 &&
                 (x < a[temp2 + 1] || x == a[temp2 + 1]) && y > a[temp2 + 1]) {
        update(temp2 + 1, 1, 0);
        ans[temp2 + 1] = 0;
      }
      a[temp1] = y;
      a[temp2] = x;
      pos[x] = pos[y];
      pos[y] = temp1;
    }
  }
  return 0;
}

```