#include <bits/stdc++.h>
using namespace std;
const int MX = 3e5 + 10, MD = 1e9 + 7;
struct node {
  long long ans, left, right;
} seg[MX * 4];
long long n, m, a[MX], ans, tmp, q, p[MX];
void build(int id = 1, int s = 1, int e = MX) {
  if (e - s == 1) {
    seg[id].left = seg[id].right = p[s];
    return;
  }
  int mid = (s + e) / 2;
  build(id * 2, s, mid);
  build(id * 2 + 1, mid, e);
  seg[id].ans = seg[id * 2].ans + seg[id * 2 + 1].ans +
                (seg[id * 2 + 1].left < seg[id * 2].right);
  seg[id].left = seg[id * 2].left;
  seg[id].right = seg[id * 2 + 1].right;
}
void upd(int x, int id = 1, int s = 1, int e = MX) {
  if (e - s == 1) {
    seg[id].left = seg[id].right = p[s];
    return;
  }
  int mid = (s + e) / 2;
  if (x < mid)
    upd(x, id * 2, s, mid);
  else
    upd(x, id * 2 + 1, mid, e);
  seg[id].ans = seg[id * 2].ans + seg[id * 2 + 1].ans +
                (seg[id * 2 + 1].left < seg[id * 2].right);
  seg[id].left = seg[id * 2].left;
  seg[id].right = seg[id * 2 + 1].right;
}
node get(int l, int r, int id = 1, int s = 1, int e = MX) {
  node ret, tmp1, tmp2;
  if (s >= r || e <= l) {
    ret.ans = 0;
    ret.left = MX;
    ret.right = -1;
    return ret;
  }
  if (s >= l && e <= r) return seg[id];
  int mid = (s + e) / 2;
  tmp1 = get(l, r, id * 2, s, mid);
  tmp2 = get(l, r, id * 2 + 1, mid, e);
  ret.ans = tmp1.ans + tmp2.ans + (tmp2.left < tmp1.right);
  ret.left = tmp1.left;
  ret.right = tmp2.right;
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]), p[a[i]] = i;
  build();
  scanf("%d", &q);
  while (q--) {
    int type, l, r;
    scanf("%d%d%d", &type, &l, &r);
    if (type == 1) printf("%lld\n", get(l, r + 1).ans + 1);
    if (type == 2) {
      swap(a[l], a[r]);
      swap(p[a[l]], p[a[r]]);
      upd(a[l]);
      upd(a[r]);
    }
  }
}

### 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>
using namespace std;
const int MX = 3e5 + 10, MD = 1e9 + 7;
struct node {
  long long ans, left, right;
} seg[MX * 4];
long long n, m, a[MX], ans, tmp, q, p[MX];
void build(int id = 1, int s = 1, int e = MX) {
  if (e - s == 1) {
    seg[id].left = seg[id].right = p[s];
    return;
  }
  int mid = (s + e) / 2;
  build(id * 2, s, mid);
  build(id * 2 + 1, mid, e);
  seg[id].ans = seg[id * 2].ans + seg[id * 2 + 1].ans +
                (seg[id * 2 + 1].left < seg[id * 2].right);
  seg[id].left = seg[id * 2].left;
  seg[id].right = seg[id * 2 + 1].right;
}
void upd(int x, int id = 1, int s = 1, int e = MX) {
  if (e - s == 1) {
    seg[id].left = seg[id].right = p[s];
    return;
  }
  int mid = (s + e) / 2;
  if (x < mid)
    upd(x, id * 2, s, mid);
  else
    upd(x, id * 2 + 1, mid, e);
  seg[id].ans = seg[id * 2].ans + seg[id * 2 + 1].ans +
                (seg[id * 2 + 1].left < seg[id * 2].right);
  seg[id].left = seg[id * 2].left;
  seg[id].right = seg[id * 2 + 1].right;
}
node get(int l, int r, int id = 1, int s = 1, int e = MX) {
  node ret, tmp1, tmp2;
  if (s >= r || e <= l) {
    ret.ans = 0;
    ret.left = MX;
    ret.right = -1;
    return ret;
  }
  if (s >= l && e <= r) return seg[id];
  int mid = (s + e) / 2;
  tmp1 = get(l, r, id * 2, s, mid);
  tmp2 = get(l, r, id * 2 + 1, mid, e);
  ret.ans = tmp1.ans + tmp2.ans + (tmp2.left < tmp1.right);
  ret.left = tmp1.left;
  ret.right = tmp2.right;
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]), p[a[i]] = i;
  build();
  scanf("%d", &q);
  while (q--) {
    int type, l, r;
    scanf("%d%d%d", &type, &l, &r);
    if (type == 1) printf("%lld\n", get(l, r + 1).ans + 1);
    if (type == 2) {
      swap(a[l], a[r]);
      swap(p[a[l]], p[a[r]]);
      upd(a[l]);
      upd(a[r]);
    }
  }
}

```