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.Arrays;
import java.util.StringTokenizer;

public class Main {

    class FenwickTree {
        public int t[];

        FenwickTree(int n) {
            t = new int[n];
        }

        void add(int index, int value) {
            while(index < t.length) {
                t[index] += value;
                index = index | (index+1);
            }
        }

        int sum(int index) {
            int res = 0;
            while (index >= 0) {
                res += t[index];
                index = (index & (index + 1)) - 1;
            }
            return res;
        }

        int sum (int left, int right) {
            return sum(right) - sum(left-1);
        }
    }

    public static void main(String[] args) {
        new Main().run();
    }

    void updateAfter(FenwickTree t, int a[], int i) {
        if (i >= a.length-1 || i < 0) return;
        if (a[i] > a[i+1]) t.add(i, 1);
    }

    void updateBefore(FenwickTree t, int a[], int i) {
        if (i >= a.length-1 || i < 0) return;
        if (a[i] > a[i+1]) t.add(i, -1);
    }

    void run() {
        int n = in.nextInt();

        int a[] = new int[n];
        int pos[] = new int[n];
        FenwickTree t = new FenwickTree(n);

        for (int i=0; i<n; i++) {
            a[i] = in.nextInt()-1;
            pos[a[i]] = i;
        }

        int prev = 0;
        for (int i=0; i<n; i++) {
            if (prev > pos[i])
                t.add(i-1, 1);
            if (i == n-1)
                t.add(i, 1);
            prev = pos[i];
        }

        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 == 1) {
                out.println(t.sum(x, y) + ((t.sum(y,y) <= 0) ? 1 : 0));
            } else {
                Arrays.stream(new int[]{a[x]-1, a[x], a[y]-1, a[y]}).distinct().forEach(value -> {updateBefore(t, pos, value); });
                int temp = a[x];
                a[x] = a[y];
                a[y] = temp;
                pos[a[x]] = x;
                pos[a[y]] = y;
                Arrays.stream(new int[]{a[x]-1, a[x], a[y]-1, a[y]}).distinct().forEach(value -> {updateAfter(t, pos, value); });
            }
        }

        out.close();
    }

    //-----------PrintWriter for faster output---------------------------------
    public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
    public static MyScanner in = new MyScanner();

    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

    }
    //--------------------------------------------------------
}
