#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
int n, m, a[N], b[N];
struct item {
  int val, le, ri;
};
item t[4 * N];
inline item mrg(item a, item b) {
  if (a.le == -1) {
    return b;
  }
  if (b.le == -1) {
    return a;
  }
  item c;
  c.val = a.val + b.val - (a.ri < b.le);
  c.le = a.le;
  c.ri = b.ri;
  return c;
}
void build(int v, int l, int r) {
  if (l == r) {
    t[v].val = 1;
    t[v].le = t[v].ri = b[l];
    return;
  }
  int mid = (r + l) >> 1;
  build(v + v, l, mid);
  build(v + v + 1, mid + 1, r);
  t[v] = mrg(t[v + v], t[v + v + 1]);
}
void update(int v, int l, int r, int pos, int val) {
  if (l == r) {
    t[v].le = t[v].ri = val;
    return;
  }
  int mid = (r + l) >> 1;
  if (pos <= mid) {
    update(v + v, l, mid, pos, val);
  } else {
    update(v + v + 1, mid + 1, r, pos, val);
  }
  t[v] = mrg(t[v + v], t[v + v + 1]);
}
item bad = {-1, -1, -1};
item get(int v, int l, int r, int tl, int tr) {
  if (l > r || l > tr || tl > r) {
    return bad;
  }
  if (tl <= l && r <= tr) {
    return t[v];
  }
  int mid = (r + l) >> 1;
  return mrg(get(v + v, l, mid, tl, tr), get(v + v + 1, mid + 1, r, tl, tr));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[a[i]] = i;
  }
  build(1, 1, n);
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int type, l, r;
    cin >> type >> l >> r;
    if (type == 1) {
      cout << get(1, 1, n, l, r).val << "\n";
    } else {
      update(1, 1, n, a[l], r);
      update(1, 1, n, a[r], l);
      swap(a[l], a[r]);
    }
  }
}

### 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;
const int N = 3e5 + 5;
int n, m, a[N], b[N];
struct item {
  int val, le, ri;
};
item t[4 * N];
inline item mrg(item a, item b) {
  if (a.le == -1) {
    return b;
  }
  if (b.le == -1) {
    return a;
  }
  item c;
  c.val = a.val + b.val - (a.ri < b.le);
  c.le = a.le;
  c.ri = b.ri;
  return c;
}
void build(int v, int l, int r) {
  if (l == r) {
    t[v].val = 1;
    t[v].le = t[v].ri = b[l];
    return;
  }
  int mid = (r + l) >> 1;
  build(v + v, l, mid);
  build(v + v + 1, mid + 1, r);
  t[v] = mrg(t[v + v], t[v + v + 1]);
}
void update(int v, int l, int r, int pos, int val) {
  if (l == r) {
    t[v].le = t[v].ri = val;
    return;
  }
  int mid = (r + l) >> 1;
  if (pos <= mid) {
    update(v + v, l, mid, pos, val);
  } else {
    update(v + v + 1, mid + 1, r, pos, val);
  }
  t[v] = mrg(t[v + v], t[v + v + 1]);
}
item bad = {-1, -1, -1};
item get(int v, int l, int r, int tl, int tr) {
  if (l > r || l > tr || tl > r) {
    return bad;
  }
  if (tl <= l && r <= tr) {
    return t[v];
  }
  int mid = (r + l) >> 1;
  return mrg(get(v + v, l, mid, tl, tr), get(v + v + 1, mid + 1, r, tl, tr));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[a[i]] = i;
  }
  build(1, 1, n);
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int type, l, r;
    cin >> type >> l >> r;
    if (type == 1) {
      cout << get(1, 1, n, l, r).val << "\n";
    } else {
      update(1, 1, n, a[l], r);
      update(1, 1, n, a[r], l);
      swap(a[l], a[r]);
    }
  }
}

```