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

import static java.lang.Math.*;

public class Solution implements Runnable {
	
	class FenwickTree {
		
		int n;
		int t [];
		
		public FenwickTree(int n) {
			this.n = n;
			t = new int [n];
		}
		
		public void set (int i, int val) {
			int value = get (i, i);
			for (; i < n; i = (i | (i + 1)))
				t[i] += (-value + val);
		}
		
		public int get (int r) {
			int result = 0;
			for (; r >= 0; r = (r & (r + 1)) - 1)
				result += t[r];
			return result;
		}
		
		public int get (int l, int r) {
			return get (r) - get (l - 1);
		}
		
	}
	
	int n;
	FenwickTree tree;
	int [] pos;
	
	void validate (int i) {
		if (i < n - 1) {
			int j = i + 1;
			if (pos[j] < pos[i]) {
				tree.set(j, 1);
			} else {
				tree.set(j, 0);
			}
		}
		if (i > 0) {
			int j = i - 1;
			if (pos[j] > pos[i]) {
				tree.set(i, 1);
			} else {
				tree.set(i, 0);
			}
		}
	}
	
	public void solve() throws Throwable {
		n = sc.nextInt();
		pos = new int [n];
		int arr [] = new int [n];
		for (int i = 0; i < n; i++) {
			arr[i] = sc.nextInt() - 1;
			pos[arr[i]] = i;
		}
		tree = new FenwickTree(n);
		tree.set(0, 1);
		for (int i = 1; i < n; i++) {
			validate(i);
		}
		int q = sc.nextInt();
		for (int j = 0; j < q; j++) {
			int type = sc.nextInt();
			if (type == 1) {
				int l = sc.nextInt() - 1;
				int r = sc.nextInt() - 1;
				int result = tree.get(l, r);
				if (tree.get(l, l) == 0) result++;
				out.println(result);
			} else {
				int posa = sc.nextInt() - 1;
				int posb = sc.nextInt() - 1;
				int a = arr[posa];
				int b = arr[posb];
				
				int t = pos[a];
				pos[a] = pos[b];
				pos[b] = t;
				
				t = arr[posa];
				arr[posa] = arr[posb];
				arr[posb] = t;
				
				validate(a);
				validate(b);
			}
		}
	}
	

	static Throwable t;

	BufferedReader in;
	FastScanner sc;
	PrintWriter out;

	@Override
	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			sc = new FastScanner(in);
			solve();
		} catch (Throwable t) {
			Solution.t = t;
		} finally {
			out.close();
		}
	}

	public static void main(String[] args) throws Throwable {
		Thread thread = new Thread(null, new Solution(), "", 1 << 26);
		thread.start();
		thread.join();
		if (Solution.t != null)
			throw t;
	}

}

class FastScanner {

	BufferedReader in;
	StringTokenizer st;

	public FastScanner(BufferedReader in) {
		this.in = in;
	}

	public String nextToken() throws Exception {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(in.readLine());
		return st.nextToken();
	}

	public int nextInt() throws Exception {
		return Integer.parseInt(nextToken());
	}

	public long nextLong() throws Exception {
		return Long.parseLong(nextToken());
	}

	public double nextDouble() throws Exception {
		return Double.parseDouble(nextToken());
	}

}