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;
struct _ {
  _() { ios_base::sync_with_stdio(0); }
} _;
template <class T>
void PV(T a, T b) {
  while (a != b) cout << *a++, cout << (a != b ? " " : "\n");
}
template <class T>
inline bool chmin(T &a, T b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
inline bool chmax(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
const int inf = 0x3f3f3f3f;
const int mod = int(1e9) + 7;
const int N = 3 * 111111;
int n, q;
int d[N];
int p[N];
int v[N];
void add(int first, int t) {
  if (first == 0) return;
  while (first <= n) {
    v[first] += t;
    first += first & (first ^ (first - 1));
  }
}
int sum(int first) {
  int r = 0;
  while (first > 0) {
    r += v[first];
    first -= first & (first ^ (first - 1));
  }
  return r;
}
int main() {
  cin >> n;
  p[0] = 0, p[n + 1] = n + 1;
  for (int i = 1; i <= n; i++) cin >> d[i], p[d[i]] = i;
  for (int i = 1; i <= n; i++)
    if (p[d[i] - 1] > i) add(d[i], 1);
  cin >> q;
  while (q--) {
    int op, l, r, t;
    cin >> op >> l >> r;
    if (op == 1) {
      cout << sum(r) - sum(l) + 1 << endl;
    } else {
      if (d[l] > d[r]) {
        t = p[d[r] - 1];
        if (l < t && t < r) add(d[r], 1);
        t = p[d[r] + 1];
        if (l < t && t < r) add(d[r] + 1, -1);
        t = p[d[l] + 1];
        if (l < t && t < r) add(d[l] + 1, 1);
        t = p[d[l] - 1];
        if (l < t && t < r) add(d[l], -1);
        if (d[l] == d[r] + 1) add(d[l], -1);
      } else if (d[l] < d[r]) {
        t = p[d[l] - 1];
        if (l < t && t < r) add(d[l], -1);
        t = p[d[l] + 1];
        if (l < t && t < r) add(d[l] + 1, 1);
        t = p[d[r] + 1];
        if (l < t && t < r) add(d[r] + 1, -1);
        t = p[d[r] - 1];
        if (l < t && t < r) add(d[r], 1);
        if (d[l] == d[r] - 1) add(d[r], 1);
      }
      swap(d[l], d[r]);
      p[d[l]] = l, p[d[r]] = r;
    }
  }
  return 0;
}
