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
import java.io.*;
import java.util.*;

public class Main implements Runnable {
 
	public void _main() throws IOException {				
		int n = nextInt();
		int[] p = new int[n];
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			p[i] = nextInt() - 1;
			a[p[i]] = i;
		}
		int[] tree = new int[n - 1];
		boolean[] was = new boolean[n - 1];
		for (int i = 0; i < n - 1; i++) {
			check(tree, n, a, was, i);			
		}
		int numQueries = nextInt();
		for (int query = 0; query < numQueries; query++) {
			int type = nextInt();
			int x = nextInt() - 1;
			int y = nextInt() - 1;
			if (type == 1) {
				out.println(1 + sum(tree, x, y - 1));
			} else {
				int npx = p[y];
				int npy = p[x];				
				p[x] = npx;
				p[y] = npy;
				a[p[x]] = x;
				a[p[y]] = y;
				check(tree, n, a, was, p[x]);
				check(tree, n, a, was, p[y]);
				check(tree, n, a, was, p[x] - 1);
				check(tree, n, a, was, p[y] - 1);
			}
		}
	}

	private void check(int[] tree, int n, int[] a, boolean[] was, int pos) {
		if (pos < 0 || pos >= n - 1) return;
		boolean f1 = was[pos];
		boolean f2 = a[pos] > a[pos + 1];
		if (f1 != f2) {
			add(tree, pos, f2 ? +1 : -1);
		}
		was[pos] = f2;
	}

	private void add(int[] tree, int pos, int val) {
		while (pos < tree.length) {
			tree[pos] += val;
			pos |= pos + 1;
		}
	}

	private int sum(int[] tree, int l, int r) {
		if (l > r) return 0;
		return sum(tree, r) - sum(tree, l - 1);
	}

	private int sum(int[] tree, int r) {
		int res = 0;
		while (r >= 0) {
			res += tree[r];
			r = (r & (r + 1)) - 1;
		}
		return res;
	}


	private BufferedReader in;
	private PrintWriter out;
	private StringTokenizer st;

	private String next() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			String rl = in.readLine();
			if (rl == null)
				return null;
			st = new StringTokenizer(rl);
		}
		return st.nextToken();
	}

	private int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	private long nextLong() throws IOException {
		return Long.parseLong(next());
	}

	private double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}

	public static void main(String[] args) {
		Locale.setDefault(Locale.UK);
		new Thread(new Main()).start();
	}

	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			//in = new BufferedReader(new FileReader("a.in"));
			//out = new PrintWriter(new FileWriter("a.out"));

			_main();

			out.close();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(202);
		}
	}

}
