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 arr[300005];
int block[605];
int blk_sz;
void update(int idx, int idy, vector<int>& pos) {
  int temp = arr[idx];
  arr[idx] = arr[idy];
  arr[idy] = temp;
  int blockNumber = idx / blk_sz;
  block[blockNumber] = 0;
  for (int i = blockNumber * blk_sz;
       i < min((int)pos.size(), (blockNumber * blk_sz) + blk_sz); i++) {
    if (i % blk_sz != 0 && pos[i - 1] > pos[i]) block[blockNumber]++;
  }
  blockNumber = idy / blk_sz;
  block[blockNumber] = 0;
  for (int i = blockNumber * blk_sz;
       i < min((int)pos.size(), (blockNumber * blk_sz) + blk_sz); i++) {
    if (i % blk_sz != 0 && pos[i - 1] > pos[i]) block[blockNumber]++;
  }
}
int query(int l, int r, vector<int>& pos, vector<int>& in) {
  int ct = 0;
  int lst = -1;
  int ql = l, qr = r;
  while (l < r and l % blk_sz != 0 and l != 0) {
    if (pos[l] < lst) ct++;
    lst = pos[l];
    l++;
  }
  while (l + blk_sz <= r) {
    ct += block[l / blk_sz];
    if ((l - 1) >= ql && (l - 1) <= qr && pos[l] < pos[l - 1]) ct++;
    l += blk_sz;
    lst = pos[l - 1];
  }
  while (l <= r) {
    if (pos[l] < lst) ct++;
    lst = pos[l];
    l++;
  }
  return ct;
}
void preprocess(vector<int>& input, vector<int>& pos, int n) {
  int blk_idx = -1;
  blk_sz = sqrt(n);
  int lst = -1;
  for (int i = 0; i < n; i++) {
    arr[i] = i;
    if (i % blk_sz == 0) {
      blk_idx++;
    } else {
      if (pos[i - 1] > pos[i]) block[blk_idx]++;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  vector<int> in;
  vector<int> pos(n);
  for (int i = 0; i < n; i++) {
    int v;
    cin >> v;
    v--;
    in.push_back(v);
    pos[v] = i;
  }
  preprocess(in, pos, n);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int p;
    cin >> p;
    int x;
    cin >> x;
    int y;
    cin >> y;
    x--;
    y--;
    if (p == 2) {
      int temp = in[x];
      in[x] = in[y];
      in[y] = temp;
      pos[in[y]] = y;
      pos[in[x]] = x;
      update(in[x], in[y], pos);
    } else {
      cout << 1 + query(x, y, pos, in) << '\n';
    }
  }
}
