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;
const int MAXN = 4e5 + 10, INF = 1e9;
int fn[MAXN], a[MAXN], b[MAXN], n;
set<pair<int, int> > s;
void upd(int p, int val) {
  for (++p; p < MAXN; p += p & (-p)) fn[p] += val;
}
int get(int p) {
  int ans = 0;
  for (; p > 0; p -= p & (-p)) ans += fn[p];
  return ans;
}
void calc(int index, int val) {
  b[index] = val;
  auto itr = s.upper_bound({index, INF});
  itr--;
  int l = itr->first, r = itr->second;
  if (index - l > 0 and b[index] <= b[index - 1]) {
    s.erase({l, r});
    s.insert({l, index - 1});
    s.insert({index, r});
    l = index;
    upd(index, 1);
  }
  if (r - index > 0 and b[index] >= b[index + 1]) {
    s.erase({l, r});
    s.insert({l, index});
    s.insert({index + 1, r});
    r = index;
    upd(index + 1, 1);
  }
  if (index == l and index > 0 and b[index] > b[index - 1]) {
    auto itr2 = s.lower_bound({index, -1});
    itr2--;
    s.erase({l, r});
    l = itr2->first;
    s.erase(itr2);
    s.insert({l, r});
    upd(index, -1);
  }
  if (index == r and index < n - 1 and b[index] < b[index + 1]) {
    auto itr2 = s.upper_bound({r, INF});
    s.erase({l, r});
    r = itr2->second;
    s.erase(itr2);
    s.insert({l, r});
    upd(index + 1, -1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
    b[a[i]] = i;
  }
  int ptr = 0;
  for (int j = 0; j < n; j++) {
    if (j > 0) {
      if (b[j] < b[j - 1]) {
        s.insert({ptr, j - 1});
        upd(ptr, 1);
        ptr = j;
      }
    }
  }
  s.insert({ptr, n - 1});
  upd(ptr, 1);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r;
      cin >> l >> r;
      cout << get(r) - get(l) + 1 << '\n';
    } else {
      int x1, x2, ind1, ind2;
      cin >> x1 >> x2;
      x1--;
      x2--;
      ind1 = a[x1];
      ind2 = a[x2];
      swap(a[x1], a[x2]);
      calc(ind1, x2);
      calc(ind2, x1);
    }
  }
  return 0;
}
