#include <bits/stdc++.h>
using namespace std;
const long long int maxn = (3e5) + 10, mod = 1e9 + 7;
long long int n, a[maxn], pnt[maxn], q;
struct node {
  long long int ans;
  long long int l, r;
  long long int res1, res2;
};
node seg[4 * maxn], ep;
node mer(node a, node b) {
  if (a.r == 0) return b;
  if (b.r == 0) return a;
  node ans;
  ans.ans = a.ans + b.ans;
  ans.l = a.l;
  ans.r = b.r;
  ans.res1 = a.res1;
  ans.res2 = b.res2;
  if (a.res2 < b.res1) ans.ans--;
  return ans;
}
void build(long long int s = 0, long long int e = n, long long int id = 1) {
  if (s == e - 1) {
    seg[id].ans = 1;
    seg[id].l = s;
    seg[id].r = e;
    seg[id].res1 = pnt[s];
    seg[id].res2 = pnt[s];
    return;
  }
  long long int mid = (s + e) >> 1;
  build(s, mid, 2 * id);
  build(mid, e, 2 * id + 1);
  seg[id] = mer(seg[2 * id], seg[2 * id + 1]);
}
void upd(long long int pntt, long long int val, long long int s = 0,
         long long int e = n, long long int id = 1) {
  if (s == e - 1) {
    pnt[s] = val;
    seg[id].ans = 1;
    seg[id].l = s;
    seg[id].r = e;
    seg[id].res1 = pnt[s];
    seg[id].res2 = pnt[s];
    return;
  }
  long long int mid = (s + e) >> 1;
  if (pntt < mid)
    upd(pntt, val, s, mid, 2 * id);
  else
    upd(pntt, val, mid, e, 2 * id + 1);
  seg[id] = mer(seg[2 * id], seg[2 * id + 1]);
}
node get(long long int l, long long int r, long long int s = 0,
         long long int e = n, long long int id = 1) {
  if (r <= s || e <= l) return ep;
  if (l <= s && e <= r) return seg[id];
  long long int mid = (s + e) >> 1;
  return mer(get(l, r, s, mid, 2 * id), get(l, r, mid, e, 2 * id + 1));
}
void deb(long long int s = 0, long long int e = n, long long int id = 1) {
  if (s == e - 1) {
    cout << seg[id].res1 << ' ';
    return;
  }
  long long int mid = (s + e) >> 1;
  deb(s, mid, 2 * id);
  deb(mid, e, 2 * id + 1);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i - 1];
    pnt[a[i - 1] - 1] = i - 1;
  }
  build();
  cin >> q;
  while (q--) {
    long long int t;
    cin >> t;
    long long int l, r;
    cin >> l >> r;
    if (t == 1)
      cout << get(l - 1, r).ans << endl;
    else {
      l--;
      r--;
      long long int res1 = a[l] - 1;
      long long int res2 = a[r] - 1;
      upd(res1, r);
      upd(res2, l);
      swap(a[l], a[r]);
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 long long int maxn = (3e5) + 10, mod = 1e9 + 7;
long long int n, a[maxn], pnt[maxn], q;
struct node {
  long long int ans;
  long long int l, r;
  long long int res1, res2;
};
node seg[4 * maxn], ep;
node mer(node a, node b) {
  if (a.r == 0) return b;
  if (b.r == 0) return a;
  node ans;
  ans.ans = a.ans + b.ans;
  ans.l = a.l;
  ans.r = b.r;
  ans.res1 = a.res1;
  ans.res2 = b.res2;
  if (a.res2 < b.res1) ans.ans--;
  return ans;
}
void build(long long int s = 0, long long int e = n, long long int id = 1) {
  if (s == e - 1) {
    seg[id].ans = 1;
    seg[id].l = s;
    seg[id].r = e;
    seg[id].res1 = pnt[s];
    seg[id].res2 = pnt[s];
    return;
  }
  long long int mid = (s + e) >> 1;
  build(s, mid, 2 * id);
  build(mid, e, 2 * id + 1);
  seg[id] = mer(seg[2 * id], seg[2 * id + 1]);
}
void upd(long long int pntt, long long int val, long long int s = 0,
         long long int e = n, long long int id = 1) {
  if (s == e - 1) {
    pnt[s] = val;
    seg[id].ans = 1;
    seg[id].l = s;
    seg[id].r = e;
    seg[id].res1 = pnt[s];
    seg[id].res2 = pnt[s];
    return;
  }
  long long int mid = (s + e) >> 1;
  if (pntt < mid)
    upd(pntt, val, s, mid, 2 * id);
  else
    upd(pntt, val, mid, e, 2 * id + 1);
  seg[id] = mer(seg[2 * id], seg[2 * id + 1]);
}
node get(long long int l, long long int r, long long int s = 0,
         long long int e = n, long long int id = 1) {
  if (r <= s || e <= l) return ep;
  if (l <= s && e <= r) return seg[id];
  long long int mid = (s + e) >> 1;
  return mer(get(l, r, s, mid, 2 * id), get(l, r, mid, e, 2 * id + 1));
}
void deb(long long int s = 0, long long int e = n, long long int id = 1) {
  if (s == e - 1) {
    cout << seg[id].res1 << ' ';
    return;
  }
  long long int mid = (s + e) >> 1;
  deb(s, mid, 2 * id);
  deb(mid, e, 2 * id + 1);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i - 1];
    pnt[a[i - 1] - 1] = i - 1;
  }
  build();
  cin >> q;
  while (q--) {
    long long int t;
    cin >> t;
    long long int l, r;
    cin >> l >> r;
    if (t == 1)
      cout << get(l - 1, r).ans << endl;
    else {
      l--;
      r--;
      long long int res1 = a[l] - 1;
      long long int res2 = a[r] - 1;
      upd(res1, r);
      upd(res2, l);
      swap(a[l], a[r]);
    }
  }
  return 0;
}

```