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 B2 {

    int calcBlock(int[] b, int x, int y) {
        int r = 1;
        for (int i = x + 1; i < y; ++i) {
            if (b[i] < b[i - 1]) ++r;
        }
        return r;
    }

    void solve() throws IOException {
        in("__std"); out("__std");

        int n = readInt();
        int[] a = new int[n];
        int[] b = new int[n];
        int k = (int) Math.sqrt(n);
        int l = n / k;
        if (n % k > 0) ++k;
        int[] c = new int[k];
        for (int i = 0; i < n; ++i) {
            a[i] = readInt() - 1;
            b[a[i]] = i;
        }
        for (int j = 0; j < k; ++j) {
            c[j] = calcBlock(b, j * l, Math.min((j + 1) * l, n));
        }
        int q = readInt();
        while (q-- > 0) {
            int p = readInt();
            int x = readInt() - 1;
            int y = readInt() - 1;
            if (p == 1) {
                int r = 1;
                if (x / l == y / l) {
                    for (int i = x + 1; i <= y; ++i) {
                        if (b[i] < b[i - 1]) ++r;
                    }
                } else {
                    for (int i = x + 1; i < (x / l + 1) * l; ++i) {
                        if (b[i] < b[i - 1]) ++r;
                    }
                    for (int j = x / l + 1; j < y / l; ++j) {
                        r += c[j];
                        if (b[j * l] > b[j * l - 1]) --r;
                    }
                    for (int i = (y / l) * l; i <= y; ++i) {
                        if (b[i] < b[i - 1]) ++r;
                    }
                }
                println(r);
            } else {
                int t = b[a[x]]; b[a[x]] = b[a[y]]; b[a[y]] = t;
                t = a[x]; a[x] = a[y]; a[y] = t;
                c[a[x] / l] = calcBlock(b, (a[x] / l) * l, Math.min((a[x] / l + 1) * l, n));
                if (a[x] / l != a[y] / l) {
                    c[a[y] / l] = calcBlock(b, (a[y] / l) * l, Math.min((a[y] / l + 1) * l, n));
                }
            }
        }

        exit();
    }

    void in(String name) throws IOException {
        if (name.equals("__std")) {
            in = new BufferedReader(new InputStreamReader(System.in));
        } else {
            in = new BufferedReader(new FileReader(name));
        }
    }

    void out(String name) throws IOException {
        if (name.equals("__std")) {
            out = new PrintWriter(System.out);
        } else {
            out = new PrintWriter(name);
        }
    }

    void exit() {
        out.close();
        System.exit(0);
    }

    char readChar() throws IOException {
        return (char) in.read();
    }

    int readInt() throws IOException {
        return Integer.parseInt(readToken());
    }

    long readLong() throws IOException {
        return Long.parseLong(readToken());
    }

    double readDouble() throws IOException {
        return Double.parseDouble(readToken());
    }

    String readLine() throws IOException {
        st = null;
        return in.readLine();
    }

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

    boolean eof() throws IOException {
        return !in.ready();
    }

    void print(String format, Object ... args) {
        out.print(new Formatter(Locale.US).format(format, args));
    }

    void println(String format, Object ... args) {
        out.println(new Formatter(Locale.US).format(format, args));
    }

    void print(Object value) {
        out.print(value);
    }

    void println(Object value) {
        out.println(value);
    }

    void println() {
        out.println();
    }

    StringTokenizer st;

    BufferedReader in;
    PrintWriter out;

    public static void main(String[] args) throws IOException {
        new B2().solve();
    }
}
