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>
class SegTree {
  std::vector<int> tree;
  int sz;

 public:
  SegTree(const std::vector<bool>& init) {
    int n = init.size();
    sz = 1;
    while (sz < n) {
      sz <<= 1;
    }
    tree.assign(2 * sz - 1, 0);
    for (int j = 0; j < n; ++j) {
      tree[j + sz - 1] = init[j];
    }
    for (int i = sz - 2; i >= 0; --i) {
      tree[i] = tree[2 * i + 1] + tree[2 * i + 2];
    }
  }
  int get_sum(int left, int right, int x = 0, int l = 0, int r = -1) const {
    if (r == -1) {
      r = sz;
    }
    if (l == left && right == r) {
      return tree[x];
    }
    int mid = (l + r) / 2, ans = 0;
    if (left < mid) {
      ans += get_sum(left, std::min(right, mid), 2 * x + 1, l, mid);
    }
    if (right > mid) {
      ans += get_sum(std::max(left, mid), right, 2 * x + 2, mid, r);
    }
    return ans;
  }
  void update(int idx, int val, int x = 0, int l = 0, int r = -1) {
    if (r == -1) {
      r = sz;
    }
    if (l + 1 == r) {
      tree[x] = val;
      return;
    }
    int mid = (l + r) / 2;
    if (idx < mid) {
      update(idx, val, 2 * x + 1, l, mid);
    } else {
      update(idx, val, 2 * x + 2, mid, r);
    }
    tree[x] = tree[2 * x + 1] + tree[2 * x + 2];
  }
};
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  int n, q, type, x, y;
  std::cin >> n;
  std::vector<int> p(n), a(n);
  for (int i = 0; i < n; ++i) {
    std::cin >> a[i];
    p[--a[i]] = i;
  }
  std::vector<bool> is_bad(n);
  for (int i = 0; i < n - 1; ++i) {
    is_bad[i] = (p[i] > p[i + 1]);
  }
  SegTree tree(is_bad);
  std::cin >> q;
  while (q--) {
    std::cin >> type >> x >> y;
    --x, --y;
    if (type == 1) {
      std::cout << 1 + tree.get_sum(x, y) << '\n';
    } else {
      int u = a[x], v = a[y];
      std::swap(a[x], a[y]);
      std::swap(p[u], p[v]);
      tree.update(u, p[u + 1] < p[u]);
      tree.update(v - 1, p[v] < p[v - 1]);
      if (u) {
        tree.update(u - 1, p[u] < p[u - 1]);
      }
      if (v < n - 1) {
        tree.update(v, p[v + 1] < p[v]);
      }
    }
  }
  return 0;
}
