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 N = 1000 * 300 + 22;
int megh[4 * N], pos[N], ary[N];
bool mark[N];
int n, q;
inline int left(int x) { return 2 * x + 1; }
inline int right(int x) { return 2 * x + 2; }
inline int update(int node, int f, int l, int ind) {
  if (l - f == 1) return megh[node] = mark[f];
  int mid = (f + l) / 2;
  if (ind < mid)
    update(left(node), f, mid, ind);
  else
    update(right(node), mid, l, ind);
  return megh[node] = megh[left(node)] + megh[right(node)];
}
inline int get_val(int node, int f, int l, int st, int en) {
  if (f == st && l == en) return megh[node];
  int mid = (f + l) / 2;
  if (en <= mid) return get_val(left(node), f, mid, st, en);
  if (st >= mid) return get_val(right(node), mid, l, st, en);
  return get_val(left(node), f, mid, st, mid) +
         get_val(right(node), mid, l, mid, en);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d", &ary[i]);
    ary[i]--;
    pos[ary[i]] = i;
  }
  for (int i = 1; i < n; i++) {
    if (pos[i] < pos[i - 1]) mark[i] = true;
    update(0, 0, n, i);
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    int ch, x, y;
    cin >> ch >> x >> y;
    x--, y--;
    if (ch == 1) {
      printf("%d\n", get_val(0, 0, n, x, y + 1) + !mark[x]);
    } else {
      swap(ary[x], ary[y]);
      pos[ary[x]] = x;
      pos[ary[y]] = y;
      x = ary[x];
      y = ary[y];
      if (x) {
        if (pos[x] < pos[x - 1])
          mark[x] = true;
        else
          mark[x] = false;
        update(0, 0, n, x);
      }
      if (y) {
        if (pos[y] < pos[y - 1])
          mark[y] = true;
        else
          mark[y] = false;
        update(0, 0, n, y);
      }
      if (x + 1 < n) {
        if (pos[x] > pos[x + 1])
          mark[x + 1] = true;
        else
          mark[x + 1] = false;
        update(0, 0, n, x + 1);
      }
      if (y + 1 < n) {
        if (pos[y] > pos[y + 1])
          mark[y + 1] = true;
        else
          mark[y + 1] = false;
        update(0, 0, n, y + 1);
      }
    }
  }
  return 0;
}
