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
#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cout << *it << " = " << a << "\n";
  err(++it, args...);
}
const int MOD = 1e9 + 7, N = 1e5 + 5, B = 32;
struct node {
  long long val = 0;
};
map<int, int> mp;
vector<node> tree;
vector<int> lazy;
int n;
bool chk(int num) {
  if (num == n or mp.find(num + 1) == mp.end()) return 1;
  return 0;
}
node comb(node a, node b) {
  node c;
  c.val = a.val + b.val;
  return c;
}
void prop(int i, int st, int en) {
  tree[i].val += lazy[i];
  if (st != en) {
    lazy[2 * i] += lazy[i];
    lazy[2 * i + 1] += lazy[i];
  }
  lazy[i] = 0;
}
void _update(int i, int st, int en, int l, int r, int val) {
  if (lazy[i]) prop(i, st, en);
  if (en < l or st > r) return;
  if (l <= st and en <= r) {
    lazy[i] = val;
    prop(i, st, en);
    return;
  }
  int mid = st + (en - st) / 2;
  _update(2 * i, st, mid, l, r, val);
  _update(2 * i + 1, mid + 1, en, l, r, val);
  tree[i] = comb(tree[2 * i], tree[2 * i + 1]);
}
void update(int l, int r, int val) { _update(1, 1, n, l, r, val); }
node _query(int i, int st, int en, int l, int r) {
  if (lazy[i]) prop(i, st, en);
  if (en < l or st > r) {
    node c;
    return c;
  }
  if (l <= st and en <= r) return tree[i];
  int mid = st + (en - st) / 2;
  return comb(_query(2 * i, st, mid, l, r),
              _query(2 * i + 1, mid + 1, en, l, r));
}
node query(int l, int r) { return _query(1, 1, n, l, r); }
void solve() {
  node dummy;
  cin >> n;
  tree.assign(4 * n, dummy);
  lazy.assign(4 * n, 0);
  vector<int> a(n + 1), giv(n + 1);
  for (int i = 1; i < n + 1; i++) cin >> a[i];
  for (int i = n; i > -1; i--) {
    giv[a[i]] = chk(a[i]);
    update(a[i], a[i], giv[a[i]]);
    mp[a[i]] = i;
  }
  int q;
  cin >> q;
  while (q--) {
    int typ, l, r;
    cin >> typ >> l >> r;
    if (typ == 1) {
      node n1 = query(l, r);
      bool fl = false;
      if (r != n and mp[r + 1] > mp[r]) fl = true;
      cout << n1.val + fl << "\n";
    } else {
      int vl = a[l], vr = a[r];
      update(vl, vl, -giv[vl]), giv[vl] = 0;
      if (vl - 1 != 0) update(vl - 1, vl - 1, -giv[vl - 1]), giv[vl - 1] = 0;
      update(vr, vr, -giv[vr]), giv[vr] = 0;
      if (vr - 1 != 0) update(vr - 1, vr - 1, -giv[vr - 1]), giv[vr - 1] = 0;
      swap(mp[vl], mp[vr]);
      swap(a[l], a[r]);
      if (vl == n or mp[vl + 1] < mp[vl]) update(vl, vl, 1), giv[vl] = 1;
      vl--;
      if (vl != 0 and (mp[vl + 1] < mp[vl])) update(vl, vl, 1), giv[vl] = 1;
      if (vr == n or mp[vr + 1] < mp[vr]) update(vr, vr, 1), giv[vr] = 1;
      vr--;
      if (vr != 0 and vr != vl + 1 and (mp[vr + 1] < mp[vr]))
        update(vr, vr, 1), giv[vr] = 1;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  solve();
}
