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 MAXN = 300001;
int read(int idx, int* tree) {
  int sum = 0;
  while (idx > 0) {
    sum += tree[idx];
    idx -= (idx & -idx);
  }
  return sum;
}
void update(int idx, int val, int* tree, int n) {
  while (idx <= n) {
    tree[idx] += val;
    idx += (idx & -idx);
  }
}
int queryGet(int x, int y, int* tree, bool* flags) {
  return read(y, tree) - read(x - 1, tree) + ((flags[y]) ? 0 : 1);
}
void queryUpd(int a, int b, int* tree, int* btop, int* ptob, bool* flags,
              int n) {
  a = ptob[a];
  b = ptob[b];
  if (a > b) {
    int tmp = a;
    a = b;
    b = tmp;
  }
  int newap = btop[b];
  int oldap = btop[a];
  int newbp = btop[a];
  int oldbp = btop[b];
  btop[a] = newap;
  btop[b] = newbp;
  ptob[newap] = a;
  ptob[newbp] = b;
  if (flags[a] && newap < oldap && a < n && btop[a + 1] > newap) {
    flags[a] = false;
    update(a, -1, tree, n);
  }
  if (flags[b] && newbp < oldbp && b < n && btop[b + 1] > newbp) {
    flags[b] = false;
    update(b, -1, tree, n);
  }
  if (!flags[a] && newap > oldap && a < n && btop[a + 1] < newap) {
    flags[a] = true;
    update(a, 1, tree, n);
  }
  if (!flags[b] && newbp > oldbp && b < n && btop[b + 1] < newbp) {
    flags[b] = true;
    update(b, 1, tree, n);
  }
  if (newap < oldap && btop[a - 1] > newap && btop[a - 1] < oldap) {
    flags[a - 1] = true;
    update(a - 1, 1, tree, n);
  }
  if (oldap < newap && btop[a - 1] < newap && btop[a - 1] > oldap) {
    flags[a - 1] = false;
    update(a - 1, -1, tree, n);
  }
  if (b != a + 1) {
    if (newbp < oldbp && btop[b - 1] > newbp && btop[b - 1] < oldbp) {
      flags[b - 1] = true;
      update(b - 1, 1, tree, n);
    }
    if (oldbp < newbp && btop[b - 1] < newbp && btop[b - 1] > oldbp) {
      flags[b - 1] = false;
      update(b - 1, -1, tree, n);
    }
  }
}
int main() {
  bool flags[MAXN] = {};
  int btop[MAXN];
  int ptob[MAXN];
  int tree[MAXN] = {};
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int tmp;
    cin >> tmp;
    flags[tmp] = true;
    flags[tmp - 1] = false;
    btop[tmp] = i + 1;
    ptob[i + 1] = tmp;
  }
  for (int i = 1; i <= n; i++) {
    if (flags[i]) update(i, 1, tree, n);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int qtype, qa, qb;
    cin >> qtype >> qa >> qb;
    if (qtype == 1)
      cout << queryGet(qa, qb, tree, flags) << endl;
    else if (qtype == 2) {
      queryUpd(qa, qb, tree, btop, ptob, flags, n);
    }
  }
}
