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 B {

	BufferedReader br;
	PrintWriter out;
	StringTokenizer st;
	boolean eof;

	static void add(int[] f, int pos, int delta) {
		for (int i = pos; i < f.length; i |= i + 1) {
			f[i] += delta;
		}
	}

	static int get(int[] f, int pos) {
		int ret = 0;
		for (int i = pos; i >= 0; i = (i & (i + 1)) - 1) {
			ret += f[i];
		}
		return ret;
	}
	
	int n;
	int[] beaver2Pos;
	int[] val, fen;

	void solve() throws IOException {
		n = nextInt();

		int[] pos2Beaver = new int[n];
		beaver2Pos = new int[n];

		for (int i = 0; i < n; i++) {
			pos2Beaver[i] = nextInt() - 1;
			beaver2Pos[pos2Beaver[i]] = i;
		}

		fen = new int[n - 1];
		val = new int[n - 1];
		for (int i = 0; i < n - 1; i++) {
			if (beaver2Pos[i] > beaver2Pos[i + 1]) {
				val[i] = 1;
				add(fen, i, 1);
			}
		}

		int q = nextInt();
		while (q-- > 0) {
			int type = nextInt();
			if (type == 1) {
				int fromBeaver = nextInt() - 1;
				int toBeaver = nextInt() - 1;
				out.println(get(fen, toBeaver - 1) - get(fen, fromBeaver - 1) + 1);
			} else {
				int pos1 = nextInt() - 1;
				int pos2 = nextInt() - 1;
				
				int beaver1 = pos2Beaver[pos1];
				int beaver2 = pos2Beaver[pos2];
				
				pos2Beaver[pos1] = beaver2;
				pos2Beaver[pos2] = beaver1;
				
				beaver2Pos[beaver1] = pos2;
				beaver2Pos[beaver2] = pos1;
				
				update(beaver1 - 1);
				update(beaver1);
				update(beaver2 - 1);
				update(beaver2);
			}
		}
	}
	
	void update(int beaver) {
		if (beaver < 0 || beaver + 1 >= n)
			return;
		int newVal = comp(beaver2Pos[beaver], beaver2Pos[beaver + 1]);
		int delta = newVal - val[beaver];
		add(fen, beaver, delta);
		val[beaver] = newVal;
	}
	
	static int comp(int a, int b) {
		if (a == b)
			throw new AssertionError();
		return a < b ? 0 : 1;
	}

	B() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new B();
	}

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				eof = true;
				return null;
			}
		}
		return st.nextToken();
	}

	String nextString() {
		try {
			return br.readLine();
		} catch (IOException e) {
			eof = true;
			return null;
		}
	}

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

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

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