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;
int n, a[301000], b[301000], where[301000];
struct Segment_Tree {
  int left, right;
  int dp, lazy;
} tree[301000 << 2];
void plant_tree(int id, int l, int r) {
  tree[id].left = l, tree[id].right = r;
  tree[id].lazy = 0;
  if (l == r) {
    tree[id].dp = b[l];
    return;
  }
  int mid = (l + r) >> 1;
  plant_tree(id << 1, l, mid);
  plant_tree(id << 1 | 1, mid + 1, r);
}
void pushdown(int id) {
  if (!tree[id].lazy) return;
  if (tree[id].left != tree[id].right) {
    tree[id << 1].lazy += tree[id].lazy;
    tree[id << 1].dp += tree[id].lazy;
    tree[id << 1 | 1].lazy += tree[id].lazy;
    tree[id << 1 | 1].dp += tree[id].lazy;
  }
  tree[id].lazy = 0;
}
void update_leaf(int id, int l, int r, int val) {
  if (tree[id].left == l && tree[id].right == r) {
    tree[id].dp += val;
    tree[id].lazy += val;
    return;
  }
  pushdown(id);
  int mid = (tree[id].left + tree[id].right) >> 1;
  if (r <= mid)
    update_leaf(id << 1, l, r, val);
  else if (mid < l)
    update_leaf(id << 1 | 1, l, r, val);
  else {
    update_leaf(id << 1, l, mid, val);
    update_leaf(id << 1 | 1, mid + 1, r, val);
  }
}
int query(int id, int pos) {
  if (tree[id].left == pos && tree[id].right == pos) {
    return tree[id].dp;
  }
  pushdown(id);
  int mid = (tree[id].left + tree[id].right) >> 1;
  if (pos <= mid)
    return query(id << 1, pos);
  else
    return query(id << 1 | 1, pos);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    where[a[i]] = i;
  }
  b[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (where[i] < where[i - 1])
      b[i] = b[i - 1] + 1;
    else
      b[i] = b[i - 1];
  }
  plant_tree(1, 1, n);
  int q;
  scanf("%d", &q);
  while (q--) {
    int ty;
    scanf("%d", &ty);
    if (ty == 1) {
      int x, y;
      scanf("%d %d", &x, &y);
      int r1 = query(1, x);
      int r2 = query(1, y);
      int ans = r2 - r1 + 1;
      printf("%d\n", ans);
    } else if (ty == 2) {
      int x, y;
      scanf("%d %d", &x, &y);
      if (a[x] + 1 == a[y]) {
        if (a[x] != 1 && where[a[x] - 1] > x && where[a[x] - 1] < y) {
          update_leaf(1, a[x], n, -1);
        }
        update_leaf(1, a[y], n, +1);
        if (a[y] != n && where[a[y] + 1] > x && where[a[y] + 1] < y) {
          update_leaf(1, a[y] + 1, n, -1);
        }
      } else if (a[x] == a[y] + 1) {
        if (a[y] != 1 && where[a[y] - 1] > x && where[a[y] - 1] < y) {
          update_leaf(1, a[y], n, +1);
        }
        update_leaf(1, a[x], n, -1);
        if (a[x] != n && where[a[x] + 1] > x && where[a[x] + 1] < y) {
          update_leaf(1, a[x] + 1, n, +1);
        }
      } else {
        if (a[x] != 1 && where[a[x] - 1] > x && where[a[x] - 1] < y) {
          update_leaf(1, a[x], n, -1);
        }
        if (a[x] != n && where[a[x] + 1] > x && where[a[x] + 1] < y) {
          update_leaf(1, a[x] + 1, n, +1);
        }
        if (a[y] != 1 && where[a[y] - 1] > x && where[a[y] - 1] < y) {
          update_leaf(1, a[y], n, +1);
        }
        if (a[y] != n && where[a[y] + 1] > x && where[a[y] + 1] < y) {
          update_leaf(1, a[y] + 1, n, -1);
        }
      }
      swap(where[a[x]], where[a[y]]);
      swap(a[x], a[y]);
    }
  }
}
