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.OutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.Reader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputStreamReader in = new InputStreamReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskB2 solver = new TaskB2();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskB2 {

    int[] t;

    int sum(int r) {
        int result = 0;
        while (r >= 0) {
            result += t[r];

            r = (r & (r + 1)) - 1;
        }
        return result;
    }

    int sum(int l, int r) {
        return sum(r) - sum(l - 1);
    }

    void inc(int i, int delta) {
        for (; i < t.length; i = (i | (i + 1))) {
            t[i] += delta;
        }
    }

    public void solve(int testNumber, InputStreamReader inSt, PrintWriter out) {
        InputReader in = new InputReader(inSt);
        int n = in.nextInt();
        int[] a = new int[n];
        int[] b = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt() - 1;
            b[a[i]] = i;
        }
        t = new int[n];
        int[] values = new int[n];
        values[0] = 1;
        inc(0, 1);
        for (int i = 0; i < n - 1; i++) {
            if (b[i] > b[i + 1]) {
                inc(i + 1, 1);
                values[i + 1] = 1;
            }
        }

        int q = in.nextInt();

        for (int i = 0; i < q; i++) {
            int p = in.nextInt();
            int x = in.nextInt() - 1;
            int y = in.nextInt() - 1;
            if (p == 2) {
                int xBobr = a[x];
                int yBobr = a[y];
                a[x] = yBobr;
                a[y] = xBobr;
                b[xBobr] = y;
                b[yBobr] = x;

                updateValues(b, values, xBobr);
                updateValues(b, values, yBobr);


            } else {
                int s = sum(x, y);
                if (values[x] == 0) {
                    s++;
                }
                out.println(s);
            }
        }
    }

    private void updateValues(int[] b, int[] values, int xBobr) {
        if (xBobr - 1 >= 0) {
            if (b[xBobr] < b[xBobr - 1]) {
                if (values[xBobr] != 1) {
                    values[xBobr] = 1;
                    inc(xBobr, 1);
                }
            } else {
                if (values[xBobr] == 1) {
                    values[xBobr] = 0;
                    inc(xBobr, -1);
                }
            }
        }
        if (xBobr + 1 < values.length) {
            if (values[xBobr + 1] == 0) {
                if (b[xBobr] > b[xBobr + 1]) {
                    values[xBobr + 1] = 1;
                    inc(xBobr + 1, 1);
                }
            } else {
                if (b[xBobr] < b[xBobr + 1]) {
                    values[xBobr + 1] = 0;
                    inc(xBobr + 1, -1);
                }
            }
        }
    }

    class InputReader {
        public BufferedReader reader;
        private String[] currentArray;
        int curPointer;

        public InputReader(InputStreamReader inputStreamReader) {
            reader = new BufferedReader(inputStreamReader);
        }

        public String next() {
            try {
                currentArray = null;
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }

        public void nextChars(char[] t) {
            try {
                currentArray = null;
                reader.read(t);
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }

        public char nextChar() {
            try {
                currentArray = null;
                return (char) reader.read();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }

        public int nextInt() {
            if ((currentArray == null) || (curPointer >= currentArray.length)) {
                try {
                    currentArray = reader.readLine().split(" ");
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
                curPointer = 0;
            }
            return Integer.parseInt(currentArray[curPointer++]);
        }

        public long nextLong() {
            if ((currentArray == null) || (curPointer >= currentArray.length)) {
                try {
                    currentArray = reader.readLine().split(" ");
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
                curPointer = 0;
            }
            return Long.parseLong(currentArray[curPointer++]);
        }

    }
}

