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 maxx = 1200010;
const double pi = acos(-1.0);
const double eps = 1e-15;
int t, n, m, op, x, y;
int ans = 0, total = 0, sum = 0, cnt = 0;
int a[maxx], b[maxx], bj[maxx];
struct node {
  int l, r, val;
} tr[maxx];
void pushup(int p) { tr[p].val = tr[p << 1].val + tr[p << 1 | 1].val; }
void build(int p, int l, int r) {
  tr[p].l = l, tr[p].r = r;
  if (l == r) {
    tr[p].val = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  pushup(p);
  return;
}
void update(int p, int c, int z) {
  int l = tr[p].l, r = tr[p].r;
  if (l == r && l == c) {
    tr[p].val = z;
    return;
  }
  int mid = (l + r) >> 1;
  if (c <= mid)
    update(p << 1, c, z);
  else
    update(p << 1 | 1, c, z);
  pushup(p);
  return;
}
int query(int p, int L, int R) {
  int l = tr[p].l, r = tr[p].r;
  if (L <= l && R >= r) return tr[p].val;
  int res = 0, mid = (l + r) >> 1;
  if (L <= mid) res += query(p << 1, L, R);
  if (R > mid) res += query(p << 1 | 1, L, R);
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]), bj[b[i]] = i;
  for (int i = 1; i <= n; i++)
    if (bj[b[i] + 1] > i) a[b[i]] = 1;
  build(1, 1, n);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      if (x != y)
        ans = y - x + 1 - query(1, x, y - 1);
      else
        ans = 1;
      printf("%d\n", ans);
    } else {
      swap(bj[b[x]], bj[b[y]]), swap(b[x], b[y]);
      if (bj[b[x] + 1] > bj[b[x]])
        update(1, b[x], 1);
      else
        update(1, b[x], 0);
      if (bj[b[y] + 1] > bj[b[y]])
        update(1, b[y], 1);
      else
        update(1, b[y], 0);
      if (b[x] - 1 != 0) {
        if (bj[b[x]] > bj[b[x] - 1])
          update(1, b[x] - 1, 1);
        else
          update(1, b[x] - 1, 0);
      }
      if (b[y] - 1 != 0) {
        if (bj[b[y]] > bj[b[y] - 1])
          update(1, b[y] - 1, 1);
        else
          update(1, b[y] - 1, 0);
      }
    }
  }
  return 0;
}
