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 V = 300100;
struct Node {
  int l, r, s;
} tr[4 * V];
void init(int id, int l, int r) {
  tr[id].l = l;
  tr[id].r = r;
  tr[id].s = 0;
  if (l == r) return;
  int mid = (l + r) / 2;
  init(id * 2, l, mid);
  init(id * 2 + 1, mid + 1, r);
}
void update(int id, int pos, int v) {
  if (tr[id].l == pos && tr[id].r == pos) {
    tr[id].s = v;
    return;
  }
  int mid = (tr[id].l + tr[id].r) / 2;
  if (pos <= mid)
    update(id * 2, pos, v);
  else
    update(id * 2 + 1, pos, v);
  tr[id].s = tr[id * 2].s + tr[id * 2 + 1].s;
}
int read(int id, int l, int r) {
  if (tr[id].l == l && tr[id].r == r) return tr[id].s;
  int mid = (tr[id].l + tr[id].r) / 2;
  ;
  if (r <= mid)
    return read(id * 2, l, r);
  else if (l > mid)
    return read(id * 2 + 1, l, r);
  else {
    int ret = 0;
    ret += read(id * 2, l, mid);
    ret += read(id * 2 + 1, mid + 1, r);
    return ret;
  }
}
int a[V], pos[V], n, Q, op, x, y;
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      pos[a[i]] = i;
    }
    init(1, 1, n);
    for (int i = 1; i < n; i++)
      if (pos[i + 1] > pos[i]) update(1, i, 1);
    scanf("%d", &Q);
    while (Q--) {
      scanf("%d%d%d", &op, &x, &y);
      if (op == 1) {
        printf("%d\n", y - x + 1 - read(1, x, y - 1));
      } else {
        swap(pos[a[x]], pos[a[y]]);
        swap(a[x], a[y]);
        if (a[x] != 1) {
          if (pos[a[x] - 1] < x)
            update(1, a[x] - 1, 1);
          else
            update(1, a[x] - 1, 0);
        }
        if (a[y] != 1) {
          if (pos[a[y] - 1] < y)
            update(1, a[y] - 1, 1);
          else
            update(1, a[y] - 1, 0);
        }
        if (a[x] != n) {
          if (x < pos[a[x] + 1])
            update(1, a[x], 1);
          else
            update(1, a[x], 0);
        }
        if (a[y] != n) {
          if (y < pos[a[y] + 1])
            update(1, a[y], 1);
          else
            update(1, a[y], 0);
        }
      }
    }
  }
  return 0;
}
