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.IOException;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author ffao
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Parser in = new Parser(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB1 solver = new TaskB1();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskB1 {
    BIT bit;
    PrintWriter out;
    int n;
    int[] p, where;

    void query(int x, int y) {
        out.println(1 + bit.read(y) - bit.read(x));
    }

    void swap(int x, int y) {
        x--; y--;
        if (x > y) {
            int t = y;
            y = x;
            x = t;
        }

        int a = p[x], b = p[y];

        if (Math.abs(a - b) == 1) {
            if (a > b) bit.update(a, -1);
            else bit.update(b, 1);
        }

        if (a != 1 && a-1 != b) {
            if (x > where[a-1] && y < where[a-1]) bit.update(a, 1);
            if (x < where[a-1] && y > where[a-1]) bit.update(a, -1);
        }

        if (b != 1 && b-1 != a) {
            if (y > where[b-1] && x < where[b-1]) bit.update(b, 1);
            if (y < where[b-1] && x > where[b-1]) bit.update(b, -1);
        }

        if (a != n && a+1 != b) {
            if (x < where[a+1] && y > where[a+1]) bit.update(a+1, 1);
            if (x > where[a+1] && y < where[a+1]) bit.update(a+1, -1);
        }

        if (b != n && b+1 != a) {
            if (y < where[b+1] && x > where[b+1]) bit.update(b+1, 1);
            if (y > where[b+1] && x < where[b+1]) bit.update(b+1, -1);
        }

        p[x] = b;
        p[y] = a;
        where[a] = y;
        where[b] = x;
    }

    public void solve(int testNumber, Parser in, PrintWriter out) {
        n = in.nextInt();
        p = new int[n];
        where = new int[n+1];
        bit = new BIT(n);
        this.out = out;

        for (int i = 0; i < n; i++) {
            p[i] = in.nextInt();
            where[p[i]] = i;
        }

        for (int i = 1; i < n; i++) {
            if (where[i+1] < where[i]) bit.update(i+1, 1);
        }

        int q = in.nextInt();
        for (int i = 0; i < q; i++) {
            int type = in.nextInt();
            int x = in.nextInt();
            int y = in.nextInt();

            if (type == 1) query(x, y);
            else swap(x, y);

            /*for (int j = 0; j < n; j++) {
                out.print(p[j] + " ");
            }
            out.println();
            for (int j = 1; j <= n; j++) {
                out.print(bit.read(j) + " ");
            }
            out.println();*/
        }
    }
}

class BIT {
    int[] tree;
    int n;

    public BIT(int n) {
        this.n = n;
        tree = new int[n+1];
    }

    public int read(int x) {
        int s = 0;
        while (x > 0) {
            s += tree[x];
            x -= x&-x;
        }
        return s;
    }

    public void update(int x, int delta) {
        while (x <= n) {
            tree[x] += delta;
            x += x&-x;
        }
    }

}

class Parser
{
    private BufferedReader din;
    private StringTokenizer tokenizer;

    public Parser(InputStream in)
    {
        din = new BufferedReader(new InputStreamReader(in));
        tokenizer = null;
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(din.readLine());
            } catch (Exception e) {
                throw new UnknownError();
            }
        }
        return tokenizer.nextToken();
    }

    public int nextInt()
    {
        return Integer.parseInt(next());
    }

    }
