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
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF331B2 extends PrintWriter {
	CF331B2() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF331B2 o = new CF331B2(); o.main(); o.flush();
	}

	int[] aa, ii;
	int[] st; int n_;
	void swap_(int i, int j) {
		int a = aa[i], b = aa[j];
		aa[i] = b; aa[j] = a;
		ii[a] = j; ii[b] = i;
	}
	void pul(int i) {
		st[i] = st[i << 1] + st[i << 1 | 1];
	}
	void build(int n) {
		n_ = n - 1;
		st = new int[n_ * 2];
		for (int i = 0; i < n_; i++)
			st[n_ + i] = ii[i] > ii[i + 1] ? 1 : 0;
		for (int i = n_ - 1; i > 0; i--)
			pul(i);
	}
	void update(int i) {
		st[i += n_] ^= 1;
		while (i > 1)
			pul(i >>= 1);
	}
	int query(int l, int r) {
		int x = 0;
		for (l += n_, r += n_; l <= r; l >>= 1, r >>= 1) {
			if ((l & 1) == 1)
				x += st[l++];
			if ((r & 1) == 0)
				x += st[r--];
		}
		return x;
	}
	void main() {
		int n = sc.nextInt();
		aa = new int[n];
		ii = new int[n];
		for (int i = 0; i < n; i++) {
			int a = sc.nextInt() - 1;
			aa[i] = a;
			ii[a] = i;
		}
		build(n);
		int q = sc.nextInt();
		while (q-- > 0) {
			int t = sc.nextInt();
			if (t == 1) {
				int a = sc.nextInt() - 1;
				int b = sc.nextInt() - 1;
				println(query(a, b - 1) + 1);
			} else {
				int i = sc.nextInt() - 1;
				int j = sc.nextInt() - 1;
				int a = aa[i];
				int b = aa[j];
				if (a > b) {
					int tmp = a; a = b; b = tmp;
					tmp = i; i = j; j = tmp;
				}
				if (a > 0 && (ii[a - 1] < i) != (ii[a - 1] < j))
					update(a - 1);
				if (a + 1 == b)
					update(a);
				else {
					if (a + 1 < n && (i < ii[a + 1]) != (j < ii[a + 1]))
						update(a);
					if (b > 0 && (ii[b - 1] < j) != (ii[b - 1] < i))
						update(b - 1);
				}
				if (b + 1 < n && (j < ii[b + 1]) != (i < ii[b + 1]))
					update(b);
				swap_(i, j);
			}
		}
	}
}
