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;
struct Tree {
  int l, r;
  int ans;
};
Tree seg[1300000];
int a[300010];
int pos[300010];
void Up(int id) {
  seg[id].ans = seg[id * 2].ans + seg[id * 2 + 1].ans;
  if (pos[seg[id * 2].r] < pos[seg[id * 2 + 1].l]) seg[id].ans--;
}
void build(int id, int l, int r) {
  seg[id].l = l;
  seg[id].r = r;
  int mid = (l + r) / 2;
  if (l == r) {
    seg[id].ans = 1;
    return;
  }
  build(id * 2, l, mid);
  build(id * 2 + 1, mid + 1, r);
  Up(id);
}
void update(int id, int pos) {
  if (seg[id].l == seg[id].r) return;
  int mid = (seg[id].l + seg[id].r) / 2;
  if (pos <= mid)
    update(id * 2, pos);
  else
    update(id * 2 + 1, pos);
  Up(id);
}
Tree query(int id, int l, int r) {
  if (seg[id].l >= l && seg[id].r <= r) return seg[id];
  int mid = (seg[id].l + seg[id].r) / 2;
  if (r <= mid)
    return query(id * 2, l, r);
  else if (l > mid)
    return query(id * 2 + 1, l, r);
  else {
    Tree ans1 = query(id * 2, l, mid);
    Tree ans2 = query(id * 2 + 1, mid + 1, r);
    Tree ans;
    ans.l = ans1.l;
    ans.r = ans2.r;
    ans.ans = ans1.ans + ans2.ans;
    if (pos[ans1.r] < pos[ans2.l]) ans.ans--;
    return ans;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  build(1, 1, n);
  int q;
  scanf("%d", &q);
  while (q--) {
    int ope, l, r;
    scanf("%d%d%d", &ope, &l, &r);
    if (ope == 1) {
      printf("%d\n", query(1, l, r).ans);
    } else {
      pos[a[l]] = r;
      update(1, a[l]);
      pos[a[r]] = l;
      update(1, a[r]);
      swap(a[l], a[r]);
    }
  }
  return 0;
}
