#include <bits/stdc++.h>
using namespace std;
const int maxN = 3e5 + 5;
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long sq(long long x) { return (x * x) % MOD; }
long long modP(long long a, long long b) {
  return (!b ? 1 : (sq(modP(a, b / 2)) * (b % 2 ? a : 1)) % MOD);
}
int n, per[maxN], inv[maxN];
struct node {
  int ans = 0, lt = -1, rt = -1;
} seg[4 * maxN];
node operator+(node A, node B) {
  if (A.lt == -1) return B;
  if (B.lt == -1) return A;
  node C;
  C.lt = A.lt;
  C.rt = B.rt;
  C.ans = A.ans + B.ans;
  if (A.rt < B.lt) C.ans--;
  return C;
}
void build(int id = 1, int s = 0, int e = n) {
  if (e - s <= 1) {
    seg[id].lt = inv[s];
    seg[id].rt = inv[s];
    seg[id].ans = 1;
    return;
  }
  int md = (s + e) / 2;
  build(2 * id, s, md);
  build(2 * id + 1, md, e);
  seg[id] = seg[2 * id] + seg[2 * id + 1];
  return;
}
void update(int p, int x, int id = 1, int s = 0, int e = n) {
  if (e - s <= 1) {
    seg[id].lt = x;
    seg[id].rt = x;
    seg[id].ans = 1;
    return;
  }
  int md = (s + e) / 2;
  if (p < md)
    update(p, x, 2 * id, s, md);
  else
    update(p, x, 2 * id + 1, md, e);
  seg[id] = seg[2 * id] + seg[2 * id + 1];
  return;
}
node get(int l, int r, int id = 1, int s = 0, int e = n) {
  if (l <= s && e <= r) return seg[id];
  if (l >= e || s >= r) return node();
  int md = (s + e) / 2;
  return get(l, r, 2 * id, s, md) + get(l, r, 2 * id + 1, md, e);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> per[i];
    per[i]--;
    inv[per[i]] = i;
  }
  build();
  int q;
  cin >> q;
  while (q--) {
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) {
      x--;
      cout << get(x, y).ans << "\n";
    } else {
      x--;
      y--;
      update(per[x], y);
      update(per[y], x);
      swap(per[x], per[y]);
      swap(inv[per[x]], inv[per[y]]);
    }
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 + 5;
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long sq(long long x) { return (x * x) % MOD; }
long long modP(long long a, long long b) {
  return (!b ? 1 : (sq(modP(a, b / 2)) * (b % 2 ? a : 1)) % MOD);
}
int n, per[maxN], inv[maxN];
struct node {
  int ans = 0, lt = -1, rt = -1;
} seg[4 * maxN];
node operator+(node A, node B) {
  if (A.lt == -1) return B;
  if (B.lt == -1) return A;
  node C;
  C.lt = A.lt;
  C.rt = B.rt;
  C.ans = A.ans + B.ans;
  if (A.rt < B.lt) C.ans--;
  return C;
}
void build(int id = 1, int s = 0, int e = n) {
  if (e - s <= 1) {
    seg[id].lt = inv[s];
    seg[id].rt = inv[s];
    seg[id].ans = 1;
    return;
  }
  int md = (s + e) / 2;
  build(2 * id, s, md);
  build(2 * id + 1, md, e);
  seg[id] = seg[2 * id] + seg[2 * id + 1];
  return;
}
void update(int p, int x, int id = 1, int s = 0, int e = n) {
  if (e - s <= 1) {
    seg[id].lt = x;
    seg[id].rt = x;
    seg[id].ans = 1;
    return;
  }
  int md = (s + e) / 2;
  if (p < md)
    update(p, x, 2 * id, s, md);
  else
    update(p, x, 2 * id + 1, md, e);
  seg[id] = seg[2 * id] + seg[2 * id + 1];
  return;
}
node get(int l, int r, int id = 1, int s = 0, int e = n) {
  if (l <= s && e <= r) return seg[id];
  if (l >= e || s >= r) return node();
  int md = (s + e) / 2;
  return get(l, r, 2 * id, s, md) + get(l, r, 2 * id + 1, md, e);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> per[i];
    per[i]--;
    inv[per[i]] = i;
  }
  build();
  int q;
  cin >> q;
  while (q--) {
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) {
      x--;
      cout << get(x, y).ans << "\n";
    } else {
      x--;
      y--;
      update(per[x], y);
      update(per[y], x);
      swap(per[x], per[y]);
      swap(inv[per[x]], inv[per[y]]);
    }
  }
  return 0;
}

```