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 = 3 * 110000;
int p[MAXN], sa[MAXN];
int n, q;
struct Node {
  int l, r;
  int sum;
  void modify(int L, int R) {
    l = L;
    r = R;
  }
  friend Node operator+(Node a, Node b) {
    Node c;
    c.modify(a.l, b.r);
    c.sum = a.sum + b.sum;
    if (sa[a.r] < sa[b.l]) c.sum--;
    return c;
  }
} T[MAXN * 4];
void build(int root, int l, int r) {
  T[root].modify(l, r);
  T[root].sum = 1;
  if (l == r) return;
  int mid = (l + r) / 2;
  build(root * 2, l, mid);
  build(root * 2 + 1, mid + 1, r);
  T[root] = T[root * 2] + T[root * 2 + 1];
}
void change(int root, int pos) {
  if (T[root].l == T[root].r) {
    T[root].sum = 1;
    return;
  }
  int mid = (T[root].l + T[root].r) / 2;
  if (pos > mid)
    change(root * 2 + 1, pos);
  else if (pos <= mid)
    change(root * 2, pos);
  T[root] = T[root * 2] + T[root * 2 + 1];
}
Node ask(int root, int l, int r) {
  if (l <= T[root].l && T[root].r <= r) {
    return T[root];
  }
  int mid = (T[root].l + T[root].r) / 2;
  if (l > mid)
    return ask(root * 2 + 1, l, r);
  else if (r <= mid)
    return ask(root * 2, l, r);
  else
    return (ask(root * 2, l, r) + ask(root * 2 + 1, l, r));
}
int brute(int l, int r) {
  int res = 0;
  for (int i = l + 1; i <= r; i++)
    if (sa[i] < sa[i - 1]) res++;
  return res + 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    sa[p[i]] = i;
  }
  build(1, 1, n);
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int type, x, y;
    scanf("%d %d %d", &type, &x, &y);
    if (type == 1) {
      printf("%d\n", ask(1, x, y).sum);
    } else {
      int xid = p[x], yid = p[y];
      swap(sa[xid], sa[yid]);
      swap(p[x], p[y]);
      change(1, xid);
      change(1, yid);
    }
  }
  return 0;
}
