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 sgt[12000003] = {0};
int a[300001];
int arr[300001];
int query(int node, int l, int r, int s, int e) {
  if (s <= l && r <= e) {
    return sgt[node];
  }
  int ret = 0;
  if ((r + l) / 2 >= s) {
    ret += query(2 * node, l, (r + l) / 2, s, e);
  }
  if ((r + l) / 2 + 1 <= e) {
    ret += query(2 * node + 1, (l + r) / 2 + 1, r, s, e);
  }
  return ret;
}
void update(int node, int l, int r, int ind, int val) {
  if (l == r) {
    sgt[node] = val;
    return;
  }
  if ((r + l) / 2 >= ind) {
    update(2 * node, l, (r + l) / 2, ind, val);
  } else {
    update(2 * node + 1, (l + r) / 2 + 1, r, ind, val);
  }
  sgt[node] = sgt[2 * node] + sgt[2 * node + 1];
}
int n;
int main() {
  scanf("%d", &n);
  int t;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &t);
    arr[i] = t;
    a[t] = i;
  }
  update(1, 1, n, 1, 1);
  for (int i = 2; i <= n; i++) {
    if (a[i] < a[i - 1]) {
      update(1, 1, n, i, 1);
    }
  }
  int q;
  scanf("%d", &q);
  int _1, _2, _3, y;
  for (int i = 0; i < q; i++) {
    scanf("%d %d %d", &_1, &_2, &_3);
    if (_1 == 1) {
      if (query(1, 1, n, _2, _2) == 0)
        y = 1;
      else
        y = 0;
      printf("%d\n", query(1, 1, n, _2, _3) + y);
    } else {
      int __2 = a[arr[_2]];
      int __3 = a[arr[_3]];
      a[arr[_2]] = __3;
      a[arr[_3]] = __2;
      int ___3 = arr[_3];
      int ___2 = arr[_2];
      int t = arr[_2];
      arr[_2] = arr[_3];
      arr[_3] = t;
      _2 = ___2;
      _3 = ___3;
      if (_3 != 1) {
        if (a[_3] > a[_3 - 1]) {
          update(1, 1, n, _3, 0);
        } else {
          update(1, 1, n, _3, 1);
        }
      }
      if (_3 != n) {
        if (a[_3] < a[_3 + 1]) {
          update(1, 1, n, _3 + 1, 0);
        } else {
          update(1, 1, n, _3 + 1, 1);
        }
      }
      if (_2 != 1) {
        if (a[_2] > a[_2 - 1]) {
          update(1, 1, n, _2, 0);
        } else {
          update(1, 1, n, _2, 1);
        }
      }
      if (_2 != n) {
        if (a[_2] < a[_2 + 1]) {
          update(1, 1, n, _2 + 1, 0);
        } else {
          update(1, 1, n, _2 + 1, 1);
        }
      }
    }
  }
}
