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 INF = int(1e9);
struct node {
  int key, prior, cnt;
  node *left, *right;
  node() {}
  node(int key)
      : key(key), prior(rand() % INF), cnt(1), left(NULL), right(NULL) {}
};
int cnt(node* v) { return (v == NULL ? 0 : v->cnt); }
void update(node*& v) {
  if (v != NULL) v->cnt = cnt(v->left) + cnt(v->right) + 1;
}
void split(node* v, node*& a, node*& b, int key) {
  if (v == NULL)
    a = b = NULL;
  else if (v->key < key) {
    split(v->right, v->right, b, key);
    a = v;
  } else {
    split(v->left, a, v->left, key);
    b = v;
  }
  update(a);
  update(b);
}
void merge(node* a, node* b, node*& v) {
  if (a == NULL)
    v = b;
  else if (b == NULL)
    v = a;
  else if (a->prior > b->prior) {
    merge(a->right, b, a->right);
    v = a;
  } else {
    merge(a, b->left, b->left);
    v = b;
  }
  update(v);
}
struct cart_tree {
  node* root;
  cart_tree() { root = new node(0); }
  void insert(int x) {
    node *a, *b, *c;
    split(root, a, b, x);
    split(b, b, c, x + 1);
    node* t = new node(x);
    merge(a, t, a);
    merge(a, c, root);
  }
  void erase(int x) {
    node *a, *b, *c;
    split(root, a, b, x);
    split(b, b, c, x + 1);
    merge(a, c, root);
  }
  int get(int x, int y) {
    node *a, *b, *c;
    split(root, a, b, x + 1);
    split(b, b, c, y + 1);
    int res = cnt(b) + 1;
    merge(a, b, a);
    merge(a, c, root);
    return res;
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  int n;
  cin >> n;
  int* A = new int[n];
  int* pos = new int[n];
  for (int i = 0; i < n; i++) {
    cin >> A[i];
    A[i]--;
    pos[A[i]] = i;
  }
  cart_tree S;
  for (int i = 0; i < n; i++) {
    if (i > 0 && pos[i] < pos[i - 1]) S.insert(i);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int p, x, y;
    cin >> p >> x >> y;
    x--, y--;
    if (p == 1)
      cout << S.get(x, y) << '\n';
    else {
      swap(pos[A[x]], pos[A[y]]);
      if (A[x] > 0 && pos[A[x]] < pos[A[x] - 1])
        S.insert(A[x]);
      else if (A[x] > 0)
        S.erase(A[x]);
      if (A[x] < n - 1 && pos[A[x]] > pos[A[x] + 1])
        S.insert(A[x] + 1);
      else if (A[x] < n - 1)
        S.erase(A[x] + 1);
      if (A[y] > 0 && pos[A[y]] < pos[A[y] - 1])
        S.insert(A[y]);
      else if (A[y] > 0)
        S.erase(A[y]);
      if (A[y] < n - 1 && pos[A[y]] > pos[A[y] + 1])
        S.insert(A[y] + 1);
      else if (A[y] < n - 1)
        S.erase(A[y] + 1);
      swap(A[x], A[y]);
    }
  }
  return 0;
}
