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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.List;
import java.util.StringTokenizer;


public class TaskB implements Runnable {

	private class Tree {
		
		public Tree(int n) {
			base = 1;
			while (base < n) {
				base <<= 1;
			}
			tree = new int[base<<1];
			Arrays.fill(tree, 0);
		}
		
		public void inc(int position) {
			position += base;
			while (position > 0) {
				++tree[position];
				position >>= 1;
			}
		}
		
		public void dec(int position) {
			position += base;
			while (position > 0) {
				--tree[position];
				position >>= 1;
			}
		}
		
		public int get(int position) {
			return tree[position+base];
		}
		
		public int get(int left, int right) {
			left += base;
			right += base;
			if (left == right) {
				return tree[left]; 
			}
			else {
				int sum = tree[left] + tree[right];
				while (left+1 < right) {
					if ((left&1) == 0) {
						sum += tree[left|1];
					}
					if ((right&1) == 1) {
						sum += tree[right^1];
					}
					left >>= 1;
					right >>= 1;
				}
				return sum;
			}
		}
		
		private int base;
		private int[] tree;
	}
	
	private void solve() {
		int q = in.readInt();
		Tree tree = new Tree(n);
		tree.inc(0);
		int position = w[0];
		for (int i = 1; i < n; ++i) {
			if (position > w[i]) {
				tree.inc(i);
			}
			position = w[i];
		}
		while (q-- > 0) {
			int p = in.readInt();
			int x = in.readInt()-1;
			int y = in.readInt()-1;
			if (p == 1) {
				int sum = tree.get(x, y);
				if (tree.get(x) == 0) {
					++sum;
				}
				out.println(sum);
			}
			else {
				int ax = a[x];
				int ay = a[y];
				if (ax != ay) {
					a[y] = ax;
					a[x] = ay;
					w[a[x]] = x;
					w[a[y]] = y;
					if (ax+1 < n && tree.get(ax+1) == 0) {
						tree.inc(ax+1);
					}
					if (tree.get(ax) == 0) {
						tree.inc(ax);
					}
					if (ay+1 < n && tree.get(ay+1) == 0) {
						tree.inc(ay+1);
					}
					if (tree.get(ay) == 0) {
						tree.inc(ay);
					}
					if (ax+1 < n && w[ax] < w[ax+1]) {
						tree.dec(ax+1);
					}
					if (ax > 0 && w[ax-1] < w[ax]) {
						tree.dec(ax);
					}
					if (ay+1 < n && w[ay] < w[ay+1] && ay+1 != ax) {
						tree.dec(ay+1);
					}
					if (ay > 0 && w[ay-1] < w[ay] && ay != ax+1) {
						tree.dec(ay);
					}
				}
			}
		}
	}
	
	private void readData() {
		n = in.readInt();
		a = new int[n];
		w = new int[n];
		for (int i = 0; i < n; ++i) {
			a[i] = in.readInt()-1;
			w[a[i]] = i;
		}
	}
	
	private int n;
	private int[] a;
	private int[] w;
		
	private void begin() {
		in = new InputReader(System.in);
		out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
	}
	
	private void end() {
		out.flush();
	}
	
	private InputReader in;
	private PrintWriter out;
	
	@Override
	public void run() {
		begin();
		readData();
		solve();		
		end();
	}

	public static void main(String[] args) {
		new TaskB().run();
	}
	
	private class InputReader {
		
	    public BufferedReader reader;
	    public StringTokenizer tokenizer;

	    public InputReader(InputStream stream) {
	        reader = new BufferedReader(new InputStreamReader(stream));
	        tokenizer = null;
	    }

	    public String next() {
	        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	                tokenizer = new StringTokenizer(reader.readLine());
	            } 
	            catch (IOException e) {
	                throw new RuntimeException(e);
	            }
	        }
	        return tokenizer.nextToken();
	    }
	    
	    public int readInt() {
	    	return Integer.parseInt(next());
	    }
	    
	    public long readLong() {
	    	return Long.parseLong(next());
	    }
	    
	    public String readString() {
	    	try {
				return reader.readLine();
			} 
	    	catch (IOException e) {				
				e.printStackTrace();
				throw new RuntimeException(e);
			}
	    }
	    
	}

}
