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 MAX_N = 300000;
int bit[MAX_N + 1], n;
int sum(int i) {
  int s = 0;
  while (i > 0) {
    s += bit[i];
    i -= i & -i;
  }
  return s;
}
void add(int i, int x) {
  if (!i) return;
  while (i <= n) {
    bit[i] += x;
    i += i & -i;
  }
}
int main() {
  cin >> n;
  int a[MAX_N + 1], pos[MAX_N + 2];
  fill(pos, pos + MAX_N + 2, 1e9);
  pos[0] = -1;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  for (int i = 1; i < n; i++) {
    if (pos[i] > pos[i + 1]) {
      add(i, 1);
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int p, x, y;
    cin >> p >> x >> y;
    if (p == 1) {
      if (x > y) {
        swap(x, y);
      }
      cout << sum(y - 1) - sum(x - 1) + 1 << endl;
    } else {
      int xb = x, yb = y;
      x = a[x];
      y = a[y];
      swap(a[xb], a[yb]);
      add(x - 1, sum(x - 2) - sum(x - 1));
      add(x, sum(x - 1) - sum(x));
      add(x + 1, sum(x) - sum(x + 1));
      add(y - 1, sum(y - 2) - sum(y - 1));
      add(y, sum(y - 1) - sum(y));
      add(y + 1, sum(y) - sum(y + 1));
      swap(pos[x], pos[y]);
      set<int> s;
      s.insert(x - 1);
      s.insert(x);
      s.insert(x + 1);
      s.insert(y - 1);
      s.insert(y);
      s.insert(y + 1);
      for (int j : s) {
        if (pos[j] > pos[j + 1]) {
          add(j, 1);
        }
      }
    }
  }
  return 0;
}
