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;
vector<long long> tree;
long long sz = 1;
long long sum(long long v, long long l, long long r, long long L, long long R) {
  if (r < L || l > R) return 0;
  if (l <= L && R <= r) return tree[v];
  return sum(v * 2 + 1, l, r, L, (L + R) / 2) +
         sum(v * 2 + 2, l, r, (L + R) / 2 + 1, R);
}
void upd(long long v) {
  if (v < 0) return;
  tree[v] = tree[2 * v + 1] + tree[2 * v + 2];
  if (v == 0) return;
  upd((v - 1) / 2);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n + 1);
  vector<long long> pos(n + 1, -1);
  vector<bool> is(n + 1);
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
    pos[a[i]] = i;
    if (a[i] == n || pos[a[i] + 1] != -1) {
      is[a[i]] = true;
    }
  }
  while (sz < n) sz *= 2;
  tree.resize(sz * 2 - 1);
  for (long long i = sz - 1; i < sz - 1 + n; ++i) {
    tree[i] = is[i - sz + 2];
  }
  for (long long i = sz - 2; i >= 0; --i) {
    tree[i] = tree[2 * i + 1] + tree[2 * i + 2];
  }
  long long q;
  cin >> q;
  while (q--) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long l, r;
      cin >> l >> r;
      r--;
      l--;
      cout << sum(0, l, r, 0, sz - 1) + (is[r + 1] == false) << '\n';
    } else {
      long long x, y;
      cin >> x >> y;
      long long left = a[x], right = a[y];
      swap(a[x], a[y]);
      pos[left] = y;
      pos[right] = x;
      is[left] = (left == n || pos[left + 1] < y);
      tree[left + sz - 2] = is[left];
      is[right] = (right == n || pos[right + 1] < x);
      tree[right + sz - 2] = is[right];
      if (left > 1)
        is[left - 1] = pos[left - 1] > pos[left],
                  tree[left + sz - 3] = is[left - 1];
      if (right > 1)
        is[right - 1] = (pos[right - 1] > pos[right]),
                   tree[right + sz - 3] = is[right - 1];
      upd((left + sz - 3) / 2);
      upd((right + sz - 3) / 2);
      if (left > 0) upd((left + sz - 4) / 2);
      if (right > 0) upd((right + sz - 4) / 2);
    }
  }
}
